Mathematics

Trigonometry: Unlocking the Secrets of Triangles

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Introduction to Trigonometry Image by Stephan Kulla – Wikimedia Commons, Wikimedia Commons, Licensed under CC0. Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles. It is a fundamental concept in mathematics and has applications in various fields such as physics, engineering, and navigation. Trigonometry helps us understand … Read more

Temporary Fix For Crankshaft Position Sensor: 9 Facts

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The Crankshaft Position Sensor is a must-have for your engine management system! 

It tells you the crankshaft’s exact spot and helps to work out ignition timing, fuel injection timing, and rotational speed. If it’s broken, watch out; it can cause misfires, stalling, engine mismanagement, poor fuel economy, and even engine overheating or failure.

Spot any warning signs?
Like a check engine light or bad engine performance? It could be a dodgy Crankshaft Position Sensor. 

temporary fix for crankshaft position sensor

To fix it, you can either replace the faulty sensor or try a temporary fix. Check the wiring harness for damaged or water-damaged circuitry, or test the voltage at the sensor terminals with a multimeter.

If you do decide to replace it, make sure you get a top-notch product from a reliable source. You may need to remove parts like the crankshaft pulley or timing belt cover to get to the sensor in the engine block. Don’t forget the electrical connector to the ECU (Engine Control Unit)!

Pro Tip: To avoid future breakdowns because of extreme temperatures, keep your vehicle’s tank full; running empty can put a strain on internal components like the Crankshaft Position Sensor. 

Time to get revved up on the critical Crankshaft Position Sensor!

What is a Crankshaft Position Sensor?

temporary fix for crankshaft position sensor

The crankshaft position sensor is essential for an engine’s functioning. 

It measures the crankshaft’s position and speed and sends this info to the engine control module (ECM). This ensures that all cylinders fire accurately, and monitors engine RPM. Keeping this part clean and replacing it regularly is key. 

Otherwise, incorrect readings can lead to bad performance, more emissions, and even damage to the engine.

Before replacing the crankshaft position sensor, make sure your vehicle needs it. Its symptoms are similar to other issues. Never change the car’s wiring or electrical systems without proper instructions.

Without the crankshaft position sensor, your car won’t know where it’s going; just like a GPS without satellites.

What does the Crankshaft Position Sensor do?

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  • The Crankshaft Position Sensor (CPS) regulates engine performance and diagnoses problems. It measures the crankshaft’s rotation speed and sends info to the ECU for fuel injection timing and spark plug firing order. The CPS also recognizes misfires and monitors mechanical functions.
  • It’s connected to a triggering wheel on the crankshaft or flywheel. This detects rotational speed and position changes of the crankshaft. It gathers precise cylinder positions and keeps the ignition system in sync.
  • However, customizations vary depending on the engine type. Some vehicles have two sensors for accuracy.
  •  Shielded wiring systems protect against interference with signal transmission.

Pro tip: Monitor wear on bearings, since faulty rotating shaft bearings produce abnormal readings from a CPS. Without it, your car’s engine is useless.

The Importance Of The Crankshaft Position Sensor.

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  • The Crankshaft Position Sensor is essential for an engine to run efficiently.
  •  Without it, misalignments could cause major breakdowns and accidents. This sensor tracks the speed and position of the crankshaft and controls the fuel injection, ignition, and other parts of the car.
  • It also protects against reverse rotation, which can cause severe damage.
  •  It monitors the signals from multiple other sensors to detect any sign of backward rotation.
  • So, don’t ignore inconsistent readings from this sensor. Get it checked regularly for proper functioning.

 If your car’s acting oddly, the cranky sensor may be to blame.

Signs Of A Faulty Crankshaft Position Sensor.

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To identify a faulty crankshaft position sensor, you need to watch for a few signs.

 This section explores the symptoms, warning signs, and trouble codes that can indicate a problem with your crankshaft position sensor. 

Knowing these sub-sections of information is essential to diagnose the issue and plan for a temporary or permanent fix, depending on the severity of the fault.

Common Symptoms of a Faulty Crankshaft Position Sensor.

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If the crankshaft position sensor is faulty, there are several signs to watch out for. These will help you to detect the problem and take action. 

Here are some of the key signals:

  • Difficulty starting.
    The engine may not start, or hesitate to start, making it hard to use the vehicle.
  • Engine stalls.
    If the engine stalls while you’re driving, it might be a faulty sensor. This can be risky on high-speed roads.
  • Loud noises.
    The engine may make loud, unusual sounds, especially when accelerating, with a damaged sensor.
  • Poor performance.
    Malfunctioning sensors can lead to slow or poor engine performance.

Also, other things can contribute to these symptoms. So, it’s best to get expert advice. A tiny issue like this can cause big damage if left untouched.

Pro Tip: Check-ups regularly can help prevent issues with the crankshaft sensor and keep your car running smoothly. If your car is showing signs, it’s got crankshaft position sensor issues.

Warning Signs.

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The Crankshaft Position Sensor is a must-have for your vehicle’s engine management system. Let’s look at the ‘Warning Signs’ that may suggest a defective crankshaft position sensor.

Symptoms: 

Signs of a faulty crankshaft position sensor could be,

  • Ignition system failure.
  • Erratic idling or acceleration.
  • Engine misfires or stalling.

Warning Lights:

  •  The dashboard’s warning lights will turn on if there’s an issue with the components.
  •  If the check engine light is on, the crankshaft position sensor could be to blame.

Error Codes: Professional mechanics can use fault scanners to look for error codes. If the crankshaft position sensor is not working, codes P0335 or P0336 are likely.

Note: These signs don’t always mean an issue with the crankshaft position sensor. Wires and cables could be malfunctioning too. So, it’s best to take your car to an expert for diagnosis.

If you suspect an issue with your Crankshaft Position Sensor, follow these steps:

  1. Take your car to a professional mechanic for an accurate diagnosis.
  2. Provide detailed information on the symptoms observed.
  3. Replace or repair any damaged parts like sensors or wires.

This will help keep your vehicle reliable and efficient.

Warning: Don’t try to diagnose your car yourself, it could cause more trouble than it’s worth!

Trouble Codes.

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  1. The ‘malfunction indicator lamp’ (MIL) in automotive engineering can signal a range of issues in vehicles. This lamp is connected to a computer chip that deciphers trouble codes. The codes tell us which problem the car has. One such code is related to crankshaft position sensors.
  2. These sensors determine the crankshaft’s position and send it to the engine control module (ECM). The ECM processes the data and makes adjustments. Faulty crankshaft position sensors can cause poor acceleration, stalling, or even prevent the car from starting.
  3. To detect issues, use a diagnostic scanner that connects to the vehicle’s OBD-II port. It will show trouble codes, which give insight into major component or system issues. 
  4. For instance, P0335 and P0336 suggest ‘crankshaft position sensor circuit malfunction’.

If you experience symptoms or get trouble codes, check your car’s sensors for wear and tear. Fix any issue quickly. Neglecting it can cause irreversible damage and increase repair costs.

We recommend visiting an experienced mechanic or taking your car to an authorized dealer if you suspect a sensor issue.

 Replacing sensors may be complex, but it can be done with DIY methods after learning relevant knowledge and certifications.

Regular maintenance and inspections are essential. Checking sensors helps improve fuel economy, prolongs vehicle lifespan, and ensures safety; giving peace of mind to drivers and passengers.

Causes Of Crankshaft Position Sensor Problems.

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To understand the causes of crankshaft position sensor problems and find a solution, dive into the Common Causes, Broken Crankshaft Position Sensors, and Faulty Wiring or Wiring Harnesses. 

This will help you identify the vital component’s signs and symptoms and why they stop working. 

You will also learn what causes poor fuel economy, engine misfires, and stalling and how to avoid or manage these issues.

Common Causes.

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Crankshaft Position Sensor Problems – Root Causes!

Is the crankshaft position sensor causing engine issues? It’s not a good look. Here’s why:

  • Wiring Issues: Wiring damage or breakage could mean signal transmission stops between the sensor and ECU.
  • Contamination: Dust and debris on the sensor or in its plug can block the signal.
  • Sensor Malfunction: Sensors lose accuracy over time, creating problems with engine speed and performance.

Remember, these causes sometimes overlap with other engine issues. So, mechanics must check all components before replacing X Sensor.

Broken Crankshaft Position Sensor.

The sensor that detects crankshaft position is a common vehicle problem. 

  • It can cause issues with fuel efficiency, emissions, and performance. 
  • This happens when the wires connected to angular position sensing are broken or damaged. It leads to problems calculating acceleration/deceleration.
  • This issue will lead to fluctuations in ignition timing vs. other sensors. Technicians use codes to detect it before it causes major engine damage. 
  • High vibration can weaken connections or malfunction the component. Misdiagnosis has caused expensive/unnecessary repairs in the past. 

So, owners must have their vehicles checked regularly to keep components functional.

Faulty Wiring or Wiring Harness.

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Inadequate cabling or wiring systems for vehicles can be a major source of trouble with crankshaft position sensors. 

  • The wiring harness helps electric signals flow, which is necessary for the sensor to work. So, if the wiring is off, the sensor will not function properly.
  • Studies have shown that many drivers of certain car models were having trouble with their crankshaft sensors. This issue was traced back to bad wiring and cabling. 
  • As a result, car makers have improved their electrical designs, since they recognize how important good wiring is.

This table shows what could happen with a wiring system failure or improper installation.

Problem Cause
Rough idling Short circuits between wires.
Intermittent start-up Loose connections or incorrect installation of connectors.
Misfire Disconnection in wire providing current.

These types of problems are usually caused by production errors or general wear over time.

If you experience any issues, it’s best to get a professional mechanic’s advice first.

 Don’t try to fix it yourself, as you might make it worse!

 Here are some temporary fixes to keep your crankshaft sensor from misbehaving:

  • Check the wiring and connectors to make sure everything is connected properly.
  • Clean the sensor to remove any dirt or debris that may be interfering with its operation.
  • Replace the sensor if it is damaged or not functioning properly.

Temporary Fixes For A Bad Crankshaft Position Sensor.

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To troubleshoot a bad crankshaft position sensor, you might need a temporary fix. This can be achieved by checking the CPS sensor with a multimeter or the CPS wiring with a test light. 

Another solution is resetting the error codes or taping the CPS connector. 

These four subsections will provide you with important information on how each solution can make your car run until you can get a permanent fix.

Checking the CPS Sensor with a Multimeter.

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Using a multimeter to inspect the crankshaft position sensor can be an efficient way of identifying issues and fixing them. 

Here’s how:

  1. Turn off the ignition. Disconnect the battery.
  2. Find the CPS sensor connector. Remove it from the sensor body.
  3. Set up the multimeter. Select DC Voltage, 20 V range. Connect to the wires.
  4. Ask an assistant or another mechanic to crank the engine. Check for voltage output from each wire.
  5. Change the setting to “Ohms“. Test each wire for resistance levels.
  6. Compare readings with manufacturer specifications.

Be careful. Repeat tests multiple times to confirm accuracy. Remember certain brands have different ranges for multimeter testing. 

Checking with an oscilloscope may give better results. For complex problems like broken components or wiring issues, a professional mechanic is recommended.

Diagnosing these issues correctly will prevent further complications, reduce expenses, keep safety standards, maintain implementation systems, prevent interruption of production time, enhance reliability, and extend performance life.

 Test the wiring with a test light for a functioning engine.

Checking CPS Wiring with a Test Light.

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For diagnosing car issues, it’s essential to check the wiring of the Crankshaft Position Sensor (CPS). Utilize a test light to accomplish this task!

  1. Take out the keys, and disconnect the wires from the CPS connector. Then attach them to the test light.
  2. Turn on the engine, but don’t start it. If the test light glows, there is a problem.
  3. Investigate if the cables are broken, or there are bad connections.

Corrosion around wiring can cause system problems. Ignoring this can result in guesswork when trying to diagnose engine performance issues.

To avoid this, regularly inspect the vehicle’s sensors. This will help detect small issues before they become huge problems. 

Resetting the error codes is like restarting life; except for relationship issues!

Resetting the Error Codes.

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If you want to get rid of the existing error codes, use a Semantic NLP variation of “Resetting the Error Codes.” 

Here’s a 5-Step Guide to resetting your vehicle’s error codes:

  1. Take an OBD scanner and read the current error codes.
  2. Write down the error codes for future reference.
  3. Disconnect the car battery and wait for 30 seconds.
  4. Reconnect the battery and start your vehicle’s engine for testing.
  5. If successful, the check engine light should not blink. Then, use the OBD scanner again to make sure there are no fault codes present.

It’s important to remember that this resetting won’t solve the underlying issue. It may only work temporarily.

To fix the problem, you need to use specialized diagnostic equipment and do more in-depth troubleshooting.

 Cartrover reports that crankshaft position sensors fail more at higher mileage.

 If duct tape can’t fix it, you’re not using enough; especially when it comes to taping the CPS connector.

Taping the CPS Connector.

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Taping the CPS Connector can be a quick and practical solution to get the car going.

 Here’s how:

  1. Wipe the connector with a dry cloth to remove dirt, debris, or moisture.
  2. Wrap the connector with electrical tape twice or thrice.
  3. Push it into place and check if it’s fixed properly.

However, taping may only give temporary relief. The tape may loosen over time or in harsh weather conditions, meaning you’ll need to reapply it.

Car From Japan shares the fact that “Crankshaft Position Sensor failures are common in both diesel and gasoline engines.” 

So, it’s time to give the crankshaft sensor a permanent fix!

Permanent Fix For A Bad Crankshaft Position Sensor.

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To permanently fix a faulty crankshaft position sensor, you need to replace it with a new one. 

Installing the new sensor can be tricky, but it’s essential for proper ignition timing and improved performance. 

Testing the new sensor ensures it’s working correctly, and your car won’t stall or misfire again.

 In this section, we will explore the subsections of replacing the crankshaft position sensor, installation of the new sensor, and testing the new sensor to provide a comprehensive guide for a permanent fix.

Replacing the Crankshaft Position Sensor.

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The Crankshaft Position Sensor can cause big problems for a vehicle’s performance. Replacing it will take care of the problem.

 Here’s a 4-step guide:

  1. Look for error codes or ask a mechanic to find the faulty sensor.
  2. Disconnect the battery cable from the negative terminal and the wiring harness from the sensor.
  3. Unscrew old sensors with pliers or wrenches and install a new one.
  4. Connect the car battery and test to ensure the sensor is working correctly.

It’s best to use an OEM replacement part from a reliable manufacturer. Aftermarket options may not be sensitive enough, causing more issues.

Letting a faulty sensor stay can cause expensive engine damage. 

A sedan owner had this problem and their mechanic warned them. The owner replaced the Crankshaft Position Sensor straight away and their vehicle’s performance improved. 

Replacing the sensor; is like a heart transplant for your car!

Installation of the New Sensor.

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To fix vehicle starting issues, a new Crankshaft Position Sensor is a must! 

Here’s a 3-step guide:

  1. Locate the old sensor under the hood.
  2. Detach wiring connectors and screws.
  3. Clean the area, install the new one, attach wiring connections, and tighten with screws.

Remember to check that the sensor’s specs match your car’s make and model. Else, compatibility problems will arise.

Clean the area with an appropriate cleaner for guaranteed installation. After fitting everything, reset possible error codes on restarting your vehicle. 

We wish the new sensor passes the test so our engine doesn’t act cranky anymore!

Testing the New Sensor.

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To double-check the accuracy of the new crankshaft sensor, a test must be done. This will show if the sensor is working right and providing reliable engine data.

The table below lists the steps to test the new sensor.

Step Action
1. Connect an oscilloscope to the Crankshaft Position Sensor (CKP) signal wire.
2. Turn on the engine and watch the waveform pattern on the oscilloscope.
3. Compare it to a known-good CKP waveform pattern to check accuracy.
4. Confirm voltage changes between high and low values of one volt or more as RPMs are increased and decreased in increments of 500 RPMs up to 5000 RPMs.

If all these values match the known-good CKP pattern, the new sensor is working properly.

Not all car makers use oscilloscopes to test CKP sensors. So, check your manufacturer’s specification tool for the right test procedure.

Pro Tip: Always stick to manufacturer instructions when testing new sensors; this helps guarantee correct readings and prevents future engine performance problems.

Fixing your car’s crankshaft sensor might be a hassle. But it’s better than the cost of a tow truck.

Frequently Asked Questions.

Q1. What is a crankshaft position sensor and why is it important?

A: A crankshaft position sensor is a device that monitors the rotational speed and position of the engine’s crankshaft. The information collected by the sensor is crucial to the engine management system, as it helps determine ignition timing, fuel injection timing, and other important parameters. The crankshaft position sensor is vital to the proper functioning of the engine, and a faulty sensor can lead to several issues including stalling, misfiring, poor fuel economy, and even engine failure.

Q2. What are the signs of a bad crankshaft position sensor?

A: The most common symptoms of a faulty crankshaft position sensor include engine misfires, stalling, rough idling, poor acceleration, and check engine light illuminated. You may also notice a decrease in engine performance and fuel economy. If you experience any of these issues, it’s important to get your vehicle inspected by a mechanic to determine the cause.

Q3. Can a bad crankshaft position sensor cause damage to the engine?

A: While a bad crankshaft position sensor won’t typically cause direct damage to the engine, it can lead to several issues that can cause harm over time. For example, a faulty sensor can cause the engine to stall or misfire, which can lead to internal combustion issues that can eventually lead to engine failure. To avoid potential damage, it’s important to address any problems with your crankshaft position sensor as soon as possible.

Q4. Is there a temporary fix for a bad crankshaft position sensor?

A: There are a few temporary fixes that can help you get your car running until you’re able to properly repair or replace the faulty sensor. One solution is to disconnect the sensor, which will cause the engine control module to use default values for timing and fuel injection. Another option is to test the sensor with a multimeter and, if it’s found to be faulty, temporarily bypass it with a jumper wire. However, it’s important to note that these temporary fixes may not work in all cases, and could lead to further issues down the line.

Q5. How much does it cost to replace a faulty crankshaft position sensor?

A: The cost of replacing a faulty crankshaft position sensor can vary depending on the make and model of your vehicle, as well as the availability of the part. On average, you can expect to pay anywhere from $100 to $300 for a new sensor and installation, though the price could be higher if additional repairs or replacement parts are needed.

Q6. Can I install a new crankshaft position sensor myself, or should I take it to a mechanic?

A: Installing a new crankshaft position sensor can be a DIY project for those with some mechanical knowledge and experience. However, it’s important to note that the installation process can be quite involved, and may require removing other components like belts or wiring harnesses. If you’re not comfortable working on your car, or if you’re unsure of your ability to complete the job safely and accurately, it’s best to take your vehicle to a qualified mechanic.

Conclusion:

The crankshaft position sensor is essential for a car’s engine. It checks the piston position and speed. A bad sensor can cause misfires, stalling, poor fuel economy, and engine failure. If you think the sensor is causing a problem, act quickly! It may not be practical or possible to replace it. But you can try a few temporary fixes. Check the wiring harness and connectors for faults or damage. Test the CKP device with a multimeter or scanner. Use a test light to test voltage levels. You can also replace the sensor with an old one or bypass it. Remember, these temporary fixes are only meant to keep your car going until you can get a new sensor. If you don’t fix the issue, it can damage other components too. So, don’t delay – seek expert help if needed.

Can Normal Distribution Be Skewed: Detailed Facts, Examples And FAQs

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Normal distribution is skewed with zero skewness, so the answer to the most common confusion can normal distribution be skewed is normal distribution is not skewed distribution as the curve of the normal distribution is symmetric without tail whose skewness is zero. The normal distribution curve is bell shaped with symmetry on the curve.

Since the skewness is lack of symmetry in the curve so if the symmetry is present in the curve there is lack of skewness.

How do you tell if the data is normally distributed?

For the data to check whether normally distributed or not just try to sketch the histogram and from the curve of the curve if the symmetry is present in the curve then the data is normally distributed, from the curve of data itself the question can normal distribution be skewed or not cleared if the concept of skewness is clear. Sketching the histogram or curve in each case is tedious or time consuming so instead of that their are number of statistical tests like Anderson-Darling statistic (AD) which are more useful to tell whether data is normally distributed or not.

The data which follows normal distribution have zero skewness in the curve and the characteristics of the curve of the skewed distribution is different without symmetry, this we will understand with the following example:

Example: Find the percent of score lies between 70 to 80 if the score of mathematics of university students are normally distributed with the mean 67 and standard deviation 9?

Can Normal Distribution Be Skewed
symmetry in the normal distribution or can normal distribution be skewed

Solution:

To find the percent of score we follow the probability for the normal distribution discussed earlier in normal distribution, so to do so first we will convert into normal variate and follow the table discussed in normal distribution to find the probability using the conversion

Z=(X-μ)/σ

we want to find the score percent between 70 and 80 so we use random variable values 70 and 80 with the given mean 67 and standard deviation 9 this gives

Z=70-67/9 = 0.333

and

Z=80-67/9 = 1.444

This we can sketch as

image 126

the above shaded area shows the region between z=0.333 and z=1.444 from the table of standard normal variate the probabilities are

P(z > 0.333)=0.3707
and
P(z > 1.444)=0.0749
so
p(0.333 < z0.333)-P(z > 1.444)=0.3707-0.0749=0.2958

so 29.58% students will score between 70 to 80 .

In the above example the skewness of the curve is zero and the curve is symmetric, to check the data is normally distributed or not we have to perform the hypothesis tests.

How do you tell if a distribution is skewed left or right?

The distribution is known to be skewed if it is right tailed or left tailed in the curve so the depending on the nature of the curve we can judge whether the distribution is positive skewed or negative skewed. The concept of skewness is discussed in detail in the articles positively and negatively skewed distribution. If the symmetry in the left side lacks the distribution is skewed left and if the symmetry lacks in the right side the distribution is skewed right. The best way to check the distribution is skewed is to check the variation in the central tendencies that is if mean<median<mode then the distribution is left skewed and if mean>median>mode then the distribution is right skewed. The geometrical representation is as follows

image 127
left skewed distribution
image 128
right skewed distribution

The measures to calculate the skewness left or right for the information given in detail in the article of skewness.

What is an acceptable skewness?

Since the skewness as earlier discussed is lack of symmetry so what range is acceptable that must be clear. The question can normal distribution is skewed arise to check whether in the normal distribution is acceptable or not and the answer of the acceptable skewness is in normal distribution because in normal distribution the skewness is zero and the distribution in which skewness is near to zero is more acceptable. So after the testing for skewness if the skewness is nearer to zero then the skewness is acceptable depending on the requirement and range for the client.

In brief the acceptable skewness is the skewness which is nearer to zero as per the requirement.

How skewed is too skewed?

The skewness is the statistical measurement to check the symmetry present in the curve of the distribution and the information and all the measures to check skewness is present or not, depending on that we can find if the distribution is far from zero then too skewed or symmetry is zero then we can say the distribution is too skewed.

How do you determine normal distribution?

To determine the distribution is normal or not we have to look the distribution have the symmetry or not if the symmetry is present and the skewness is zero then the distribution is normal distribution, the detail methods and techniques were already discussed in detail in normal distribution

Do outliers skew data?

In the distribution data if any data follow unusual way and very far or away from the usual data that is known as outlier and in most of the cases the outliers are responsible for the skewness of the distribution and because of the unusual nature of outliers the distribution have skewness, so we can say that in the distribution the outliers skew data. The outliers in all cases will not skew data they skewed data only if they also follow the systematic sequence in continuous distribution to give left or right tailed curve.

In the previous articles the detail discussion of normal distribution and skewed distribution discussed.

Hermite Polynomial: 9 Complete Quick Facts

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image 145

  The Hermite polynomial is widely occurred in applications as an orthogonal function. Hermite polynomial is the series solution of Hermite differential equation.

Hermite’s Equation

    The differential equation of second order with specific coefficients as

d2y/dx2 – 2x dy/dx + 2xy = 0

is known as Hermite’s equation, by solving this differential equation we will get the polynomial which is Hermite Polynomial.

Let us find the solution of the equation

d2y/dx2 – 2x dy/dx + 2ny = 0

with the help of series solution of differential equation

101 1

now substituting all these values in the Hermite’s equation we have

image 136

This equation satisfies for the value of k=0 and as we assumed the value of k will not be negative, now for the lowest degree term xm-2 take k=0 in the first equation as the second gives negative value, so the coefficient xm-2 is

a0m (m-1)=0 ⇒ m=0,m=1

as a0 ≠ 0

now in the same way equating the coefficient of xm-1 from the second summation

104

and equating the coefficients of xm+k to zero,

ak+2(m+k+2)(m+k+1)-2ak(m+k-n) = 0

we can write it as

ak+2 = 2(m+k-n)/(m+k+2)(m+k+1) ak

if m=0

ak+2 = 2(k-n)/(k+2)(k+1) ak

if m=1

ak+2 = 2(k+1-n)/(k+3)(k+2) ak

for these two cases now we discuss the cases for k

When $m=0, ak+2= 2(k-n)/ (k+2)(k+1)} ak$

If, $k=0 a2 =-2 n/2 a0=-n a0$

$k=1, a3=2(1-n)/6 a1 =-2(n-1)/3 ! a1$

If $k=2, a4 =2(2-n)/12 a2 =2 (2-n)/12 (-n a0) =22 n(n-2)/4 ! a0$

108

so far m=0 we have two conditions when a1=0, then a3=a5=a7=….=a2r+1=0 and when a1 is not zero then

image 140

by following this put the values of a0,a1,a2,a3,a4 and a5 we have

image 141

and for m=1 a1=0 by putting k=0,1,2,3,….. we get

ak+2 = 2(k+1-n)/(k+3)(k+2)ak

image 142

so the solution will be

image 143

so the complete solution is

image 144

where A and B are the arbitrary constants

Hermite Polynomial

   The Hermite’s equation solution is of the form y(x)=Ay1(x)+By2(x) where y1(x) and y2(x) are the series terms as discussed above,

image 145
image 146

one of these series end if n is non negative integer if n is even y1 terminates otherwise y2 if n is odd, and we can easily verify that for n=0,1,2,3,4…….. these polynomials are

1,x,1-2x2, x-2/3 x3, 1-4x2+4/3x4, x-4/3x3+ 4/15x5

so we can say here that the solution of Hermite’s equation are constant multiple of these polynomials and the terms containing highest power of x is of the form 2nxn denoted by Hn(x) is known as Hermite polynomial

Generating function of Hermite polynomial

Hermite polynomial usually defined with the help of relation using generating function

image 150
image 149

[n/2] is the greatest integer less than or equal to n/2 so it follows the value of Hn(x) as

image 151
image 152

this shows that Hn(x) is a polynomial of degree n in x and

Hn(x) = 2nxn + πn-2 (x)

where πn-2 (x) is the polynomial of degree n-2 in x, and it will be even function of x for even value of n and odd function of x for odd value of n, so

Hn(-x) = (-1)n Hn(x)

some of the starting Hermite polynomials are

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 – 2

H3(x) = 8x3-12

H4(x) = 16x4 – 48x2+12

H5(x) = 32x2 – 160x3+120x

Generating function of Hermite polynomial by Rodrigue Formula

Hermite Polynomial can also be defined with the help of Rodrigue formula using generating function

image 153

since the relation of generating function

image 154

  Using the Maclaurin’s theorem, we have

image 155

or

by putting z=x-t and

for t=0,so z=x gives

this we can show in another way as

differentiating

with respect to t gives

taking limit t tends to zero

now differentiating with respect to x

taking limit t tends to zero

from these two expressions we can write

in the same way we can write

 differentiating n times put t=0, we get

from these values we can write

from these we can get the values

Example on Hermite Polynomial           

  1. Find the ordinary polynomial of

Solution: using the Hermite polynomial definition and the relations we have

2. Find the Hermite polynomial of the ordinary polynomial

Solution: The given equation we can convert to Hermite as

and from this equation equating the same powers coefficient

hence the Hermite polynomial will be

Orthogonality of Hermite Polynomial | Orthogonal property of Hermite Polynomial

The important characteristic for Hermite polynomial is its orthogonality which states that

To prove this orthogonality let us recall that

which is the generating function for the Hermite polynomial and we know

so multiplying these two equations we will get

multiplying and integrating within infinite limits

and since

so

using this value in above expression we have

which gives

now equate the coefficients on both the sides

which shows the orthogonal property of Hermite polynomial.

  The result of orthogonal property of Hermite polynomial can be shown in another way by considering the recurrence relation

Example on orthogonality of Hermite Polynomial

1.Evaluate the integral

Solution: By using the property of orthogonality of hermite polynomial

since the values here are m=3 and n=2 so

2. Evaluate the integral

Solution: Using the orthogonality property of Hermite polynomial we can write

Recurrence relations of Hermite polynomial

The value of Hermite polynomial can be easily find out by the recurrence relations

Hermite polynomial
Hermite polynomial recurrence relations

These relations can easily obtained with the help of definition and properties.

Proofs:1. We know the Hermite equation

y”-2xy’+2ny = 0

and the relation

image 174

by taking differentiation with respect to x partially we can write it as

image 175

from these two equations

image 176
image 177

now replace n by n-1

image 178
image 179

by equating the coefficient of tn

image 180
image 181

so the required result is

image 182

2. In the similar way differentiating partially with respect to t the equation

image 183

we get

image 184
image 185

n=0 will be vanished so by putting this value of e

image 186
image 187

now equating the coefficients of tn

image 188

thus

image 189

3. To prove this result we will eliminate Hn-1 from

image 190

and

image 191

so we get

image 192

thus we can write the result

image 193

4. To prove this result we differentiate

image 194

we get the relation

image 195

substituting the value

image 196

and replacing n by n+1

image 197

which gives

image 173

Examples on Recurrence relations of Hermite polynomial

1.Show that

H2n(0) = (-1)n. 22n (1/2)n

Solution:

To show the result we have

image 172

H2n(x) =

taking x=0 here we get

image 171

2. Show that

H’2n+1(0) = (-1)n 22n+1 (3/2)2

Solution:

Since from the recurrence relation

H’n(x) = 2nHn-1(X)

here replace n by 2n+1 so

H’2n-1(x) = 2(2n+1) H2n(x)

taking x=0

image 170

3. Find the value of

H2n+1(0)

Solution

Since we know

image 169

use x=0 here

H2n-1(0) = 0

4. Find the value of H’2n(0).

Solution :

we have the recurrence relation

H’n(x) = 2nHn-1(x)

here replace n by 2n

H’2n(x) = =2(2n)H2n-1(x)

put x=0

H’2n(0) = (4n)H2n-1(0) = 4n*0=0

5. Show the following result

image 168

Solution :

Using the recurrence relation

H’n(x) = 2nHn-1 (x)

so

image 167

and

d3/dx3 {Hn(x)} = 23n(n-1)(n-2)Hn-3(x)

differentiating this m times

image 166

which gives

image 165

6. Show that

Hn(-x) = (-1)n Hn(x)

Solution :

we can write

image 163
image 164

from the coefficient of tn we have

image 162

and for -x

image 161

7. Evaluate the integral and show

Solution : For solving this integral use integration parts as

image 160

Now differentiation under the Integral sign differentiate with

respect to x

image 159

using

H’n(x) = 2nHn-1 (x)

and

H’m(x) = 2mHm-1 (x)

we have

image 157

and since

???? n,m-1 = ????n+1, m

so the value of integral will be

image 156

Conclusion:

The specific polynomial which frequently occurs in application is Hermite polynomial, so the basic definition, generating function , recurrence relations and examples related to Hermite Polynomial were discussed in brief here   , if you require further reading go through

https://en.wikipedia.org/wiki/Hermite_polynomials

For more post on mathematics, please follow our Mathematics page

2D Coordinate Geometry: 11 Important Facts

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Screenshot 59 300x212 1

Locus in 2D Coordinate Geometry

Locus is a Latin word. It is derived from the word ‘Place’ or ‘Location’. The Plural of locus is Loci.

Definition of Locus:

In Geometry, ‘Locus’ is a set of points which satisfy one or more specified conditions of a figure or shape. In modern mathematics, the location or the path on which a point moves on the plane satisfying given geometrical conditions, is called locus of the point.

Locus is defined for line, line segment and the regular or irregular curved shapes except the shapes having vertex or angles inside them in Geometry. https://en.wikipedia.org/wiki/Coordinate_system

Examples on Locus:

lines, circles, ellipse, parabola, hyperbola etc. all these geometrical shapes are defined by the locus of points.

Equation of the Locus:

The algebraic form of the geometrical properties or conditions which are satisfied by the coordinates of all the points on Locus, is known as the equation of the locus of those points.

Method of Obtaining the Equation of the Locus:

To find the equation of the locus of a moving point on a plane, follow the process described below

(i) First, assume the coordinates of a moving point on a plane be (h,k).

(ii) Second, derive a algebraic equation with h and k from the given geometrical conditions or properties.

(iii) Third, replace h and k by x and y respectively in the above said equation. Now this equation is called the the equation of the locus of the moving point on the plane. (x,y) is the current coordinates of the moving point and the equation of the locus must always be derived in the form of x and y i.e. current coordinates.

Here are some examples to make the conception clear about locus.

4+different types of solved problems on Locus:

Problem 1: If P be any point on the XY-plane which is equidistant from two given points A(3,2) and B(2,-1) on the same plane, then find the locus and the equation of locus of the point P with graph.

Solution: 

Locus
Graphical representation

Assume that the coordinates of any point on the locus of P on XY-plane are (h, k).

Since, P is equidistant from A and B, we can write

The distance of P from A=The distance of P from B

Or, |PA|=|PB|

lagrida latex editor 51
lagrida latex editor 46

Or, (h2 -6h+9+k2 -4k+4) = (h2 -4h+4+k2 +2k+1)——– taking square to both sides.

Or, h2 -6h+13+k2 -4k -h2+4h-5-k2 -2k = 0

Or, -2h -6k+8 = 0

Or, h+3k -4 = 0

Or, h+3k = 4 ——– (1)

This is a first degree equation of h and k.

Now if h and k are replaced by x and y then the equation (1) becomes the first degree equation of x and y in the form of x + 3y = 4 which represents a straight line.

Therefore, the locus of the point P(h, k) on XY-plane is a straight line and the equation of the locus is x + 3y = 4 . (Ans.)

Problem 2: If a point R moves on the XY-plane in such way that RA : RB = 3:2 where the coordinates of the points A and B are (-5,3) and (2,4) respectively on the same plane, then find the locus of the point R.

What type of curve does the equation of the locus of R indicate?

Solution: Lets assume that the coordinates of any point on the locus of given point R on XY-plane be (m, n).

Asper given condition RA : RB = 3:2,

we have,

(The distance of R from A) / (The distance of R from B) = 3/2

lagrida latex editor 47

Or, (m2 +10m+34+n2 -6n) / (m2 -4m+n2 -8n+20) =9/4 ———– taking square to both sides.

Or, 4(m2 +10m+34+n2 -6n) = 9(m2 -4m+n2 -8n+20)

Or, 4m2 +40m+136+4n2 -24n = 9m2 -36m+9n2 -72n+180)

Or, 4m2 +40m+136+4n2 -24n – 9m2 +36m-9n2 +72n-180 = 0

Or, -5m2 +76m-5n2+48n-44 = 0

Or, 5(m2+n2)-76m+48n+44 = 0 ———-(1)

This is a second degree equation of m and n .

Now if m and n are replaced by x and y, the equation (1) becomes the second degree equation of x and y in the form of 5(x2+y2)-76x+48y+44 = 0 where the coefficients of x2 and y2 are same and the coefficient of xy is zero. This equation represents a circle.

Therefore, the locus of the point R(m, n) on XY-plane is a circle and the equation of the locus is

5(x2+y2)-76x+48y+44 = 0 (Ans.)

Problem 3: For all values of (θ,aCosθ,bSinθ) are the coordinates a point P which moves on the XY plane. Find the equation of locus of P.

Solution: lets (h, k) be the coordinates of any point lying on the locus of P on XY-plane.

Then asper the question, we can say

h= a Cosθ

Or, h/a = Cosθ —————(1)

And k = b Sinθ

Or, k/b = Sinθ —————(2)

Now taking square of both the equations (1) and (2) and then adding, we have the equation

h2/a2 + k2/b2 =Cos2θ + Sin2θ

Or, h2/a2 + k2/b2 = 1 (Since Cos2θ + Sin2θ =1 in trigonometry)

Therefore the equation of locus of the point P is x2/a2 + y2/b2 = 1 . (Ans.)

Problem 4 : Find the equation of locus of a point Q, moving on the XY-plane, if the coordinates of Q are

lagrida latex editor 1 1

where u is the variable parameter.

Solution : Let the coordinates of any point on the locus of given point Q while moving on XY-plane be (h, k).

Then, h = lagrida latex editor 3and k = lagrida latex editor 2

i.e. h(3u+2) = 7u-2 and k(u-1) = 4u+5

i.e. (3h-7)u = -2h-2 and (k-4)u = 5+k

i.e. u =lagrida latex editor 4 —————(1)

and u = lagrida latex editor 5 —————(2)

Now equating the equations (1) and (2) , we get, lagrida latex editor 6

Or, (-2h-2)(k-4) = (3h-7)(5+k)

Or, -2hk+8h-2k+8 = 15h+3hk-35-7k

Or, -2hk+8h-2k-15h-3hk+7k = -35-8

Or, -5hk-7h+5k = -43

Or, 5hk+7h-5k = 43

Therefore, the equation of the locus of Q is 5xy+7x-5y = 43.

More examples on Locus with answers for practice by your own:

Problems 5: If θ be a variables and u be a constant, then find the equation of locus of the point of intersection of the two straight lines x Cosθ + y Sinθ = u and x Sinθ- y Cosθ = u. ( Ans. x2+y2 =2u2 )

Problems 6: Find the equation of locus of the middle point of the line segment of the straight line x Sinθ + y Cosθ = t between the axes. ( Ans. 1/x2+1/y2 =4/t2 )

Problems 7: If a point P is moving in such way on the XY-plane that the area of the triangle made by the point with two points (2,-1) and (3,4). ( Ans. 5x-y=11)

Basic Examples on the Formulae “Centroid of a Triangle”  in 2D Coordinate Geometry

Centroid: The three medians of a triangle always intersect at a point, located in the interior area of the triangle and divides the median at the ratio 2:1 from any vertex to the midpoint of the opposite side. This point is called the centroid of the triangle.   

Problems 1:  Find the centroid of the triangle with vertices (-1,0), (0,4) and (5,0).

Solution:  We already know,

                                             If  A(x1,y1) , B(x2,y2) and C(x3,y3) be the vertices of a Triangle and G(x, y) be the centroid of the triangle, then Coordinates of G are

lagrida latex editor 7

and

lagrida latex editor 8 1

Using this formula we have , 

(x1,y1) ≌(-1,0) i.e. x1= -1, y1=0 ;

(x2,y2) ≌(0,4) i.e.   x2=0, y2=4 and

(x3,y3) ≌(5,0)  i.e.   x3=5, y3=0

(See formulae chart)

Screenshot 17
Graphical Representation

So, the x-coordinate of the centroid G,   lagrida latex editor 9

i.e. lagrida latex editor 10

i.e. x=4/3

                  and 

the y-coordinate of the centroid G,  lagrida latex editor 11

i.e lagrida latex editor 12

i.e y=4/3

Therefore, the coordinates of the centroid of the given triangle is lagrida latex editor 13 . (Ans)

More answered problems are given below for further practice using the procedure described in above problem 1 :-

Problems 2: Find the coordinates of the centroid of the triangle with vertices at the points (-3,-1), (-1,3)) and (1,1).

Ans. (-1,1)

Problems 3: What is the x-coordinate of the centroid of the triangle with vertices (5,2), (10,4) and (6,-1) ?

Ans. 7 

Problems 4: Three vertices of a triangle are (5,9), (2,15) and (11,12).Find the centroid of this triangle.

Ans. (6,12)

Shifting of Origin / Translation of Axes- 2D Co-ordinate Geometry

Shifting of Origin means to shift the Origin to a new point keeping the orientation of the axes unchanged i.e the new axes remain parallel to the original axes in the same plane. By this translation of axes or shifting of origin process many problems on algebraic equation of a geometric shape are simplified and solved easily.

The formula of ” Shifting of Origin” or “Translation of Axes” are described below with graphical representation.

Formula:

If O be the origin ,P(x,y) be any point in the XY plane and O be  shifted to another point O′(a,b) against which the coordinates of the point P become (x1,y1) in the same plane with new axes X1Y1  ,Then New Coordinates of P are

x1 = x- a

y1 = y- b

Graphical representation for clarification: Follow the graphs

Screenshot 45
Screenshot 46

Few solved Problems on the formula of ‘Shifting of Origin’ :

Problem-1 : If there are two points (3,1) and (5,4) in the same plane and the origin is shifted to the point (3,1) keeping the new axes parallel to the original axes, then find the co-ordinates of the point (5,4) in respect with the new origin and axes.

Solution: Comparing with the formula of ‘Shifting of Origin’ described above , we have new Origin, O′(a, b) ≌ (3,1) i.e. a=3 , b=1 and the required point P, (x, y) ≌ (5,4) i.e. x=5 , y=4

Screenshot 52

Now if (x1,y1) be the new coordinates of the point P(5,4) ,then asper formula x1 = x-a and y1 =y-b,

we get, x1 = 5-3 and y1 =4-1

i.e. x1 = 2 and y1 =3

Therefore, the required new coordinates of the point (5,4) is (2,3) . (Ans.)

Problem-2 : After shifting the Origin to a point in the same plane ,remaining the axes parallel to each other ,the coordinates of a point (5,-4) become (4,-5).Find the Coordinates of new Origin.

Solution: Here using the formula of ‘Shifting the Origin’ or ‘Translation of Axes’ , we can say the coordinates of the point P with respect to old and new Origin and axes respectively are (x, y) ≌ (5,-4) i.e. x=5 , y= -4 and (x1,y1) ≌ (4,-5) i.e.  x1= 4, y1= -5

Screenshot 50

Now we have to find the coordinates of the new Origin O′(a, b) i.e. a=?, b=?

Asper formula,

x1 = x- a

y1 = y- b

i.e. a=x-x1 and b=y-y1

Or, a=5-4 and b= -4-(-5)

Or, a=1 and b= -4+5

Or, a=1 and b= 1

Therefore, O'(1,1) be the new Origin i.e. the coordinates of the new Origin are (1,1). (Ans.)

Basic Examples on the Formulae “Collinearity of points (three points)” in 2D Coordinate Geometry

Problems 1:  Check whether the points (1,0), (0,0) and (-1,0) are collinear or not.

Solution:  We already know,

                                            If  A(x1,y1) , B(x2,y2) and C(x3,y3) be any three collinear points, then the area of the triangle made by them must be zero i.e the area of the triangle is ½[x1 (y2– y3) + x2 (y3– y1) + x3 (y1-y2)] =0

(See formulae chart)

Using this formula we have ,

(x1,y1) ≌(-1,0) i.e.   x1= -1, y1= 0   ;

(x2,y2) ≌(0,0)  i.e.   x2= 0, y2= 0;

(x3,y3) ≌(1,0)  i.e.    x3= 1, y3= 0

Screenshot 14
Graphical Representation

So, the area of the triangle is = |½[x1 (y2–  y3) + x2 (y3–  y1) + x3 (y1-y2)]| i.e.

(L.H.S) = |½[-1 (0-0) + 0 (0-0) + 1 (0-0)]|

= |½[(- 1)x0 + 0x0 + 1×0]|

= |½[0 + 0 + 0]|

= |½ x 0|

= 0  (R.H.S)

Therefore, the area  of the triangle made by those given points become zero which means they are lying on the same line.

Therefore, the given points are collinear points. (Ans)

More answered problems are given below for further practice using the procedure described in the above problem 1 :-

Problems 2: Check whether the points (-1,-1), (0,0) and (1,1) are  collinear or not.

Ans. Yes

Problems 3: Is it possible to draw one line through three points (-3,2), (5,-3) and (2,2) ?

Ans.No

Problems 4: Check whether the points (1,2), (3,2) and (-5,2),connected by lines, can form a triangle in the coordinate plane.

Ans. No

______________________________

Basic Examples on the Formulae “Incenter of a Triangle” in 2D Coordinate Geometry

Incenter:It is the center of the triangle’s largest incircle which fits inside the triangle.It is also the point of intersection of the three bisectors of the interior angles of the triangle.

Problems 1: The vertices of a triangle with sides  are (-2,0), (0,5) and (6,0) respectively. Find the incenter of the triangle.

Solution: We already know,

If  A(x1,y1) , B(x2,y2) and C(x3,y3) be the vertices, BC=a, CA=b and AB=c , G′(x,y) be the incentre of the triangle,

The co-ordinates of G′ are

lagrida latex editor 14 1

and         

lagrida latex editor 15 1

(See formulae chart)

Screenshot 56

Asper the formula we have,

(x1,y1) ≌(-4,0) i.e.  x1= -4, y1=0 ;

(x2,y2) ≌(0,3) i.e.  x2=0, y2=3 ;

(x3,y3) ≌(0,0)  i.e.   x3=0, y3=0

We have now,

a= √ [(x2-x1)2+(y2-y1)2 ]

Or, a= √ [(0+4)2+(3-0)2 ]

Or, a= √ [(4)2+(3)2 ]

Or, a= √ (16+9)

Or, a= √25

Or, a = 5 ——————(1)

b=√ [(x1-x3)2+(y1-y3)2 ]

Or, b= √ [(-4-0)2+(0-0)2 ]

Or, b= √ [(-4)2+(0)2 ]

Or, b= √ (16+0)

Or, b= √16

Or, b= 4 ——————–(2)

c= √ [(x3-x2)2+(y3-y2)2 ]

Or, c= √ [(0-0)2+(0-3)2 ]

Or, c= √ [(0)2+(-3)2 ]

Or, c= √ (0+9)

Or, c= √9

Or, c= 3 ——————–(3)

and ax1+ bx2 + cx3 = (5 X (-4)) + (4 X 0) + (3 X 6 )

= -20+0+18

Or, ax1+ bx2 + cx3 = -2 ——————-(4)

ay1+ by2+ cy3 = (5 X 0) + (4 X 3) + (3 X 0)

= 0+12+0

Or, ay1+ by2+ cy3 = 12 ——————–(5)

a+b+c = 5+4+3

Or, a+b+c = 12 ——————(6)

Using the above equations (1), (2), (3), (4), (5) and (6) we can calculate the value of x and y from

lagrida latex editor 16 1

Or, x = -2/12

Or, x = -1/6

and

lagrida latex editor 17 1

Or, y = 12/12

Or, y = 1

Therefore the required coordinates of the incenter of the given triangle are (-1/6 , 1). (Ans.)

More answered problems are given below for further practice using the procedure described in above problem 1 :-

Problems 2: Find the coordinates of the incenter of the triangle with vertices at the points (-3,-1), (-1,3)) and (1,1).

Problems 3: What is the x-coordinate of the incenter of the triangle with vertices (0,2), (0,0) and (0,-1) ?

Problems 4: Three vertices of a triangle are (1,1), (2,2) and (3,3). Find the incenter of this triangle.

13 Facts On Chebyshev’s Inequality & Central Limit Theorem

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In the probability theory the Chebyshev’s Inequality & central limit theorem deal with the situations where we want to find the probability distribution of sum of large numbers of random variables in approximately normal condition, Before looking the limit theorems we see some of the inequalities, which provides the bounds for the probabilities if the mean and variance is known.

Markov’s inequality

The Markov’s inequality for the random variable X which takes only positive value for a>0 is

gif

to prove this for a>0 consider

Since

gif

now taking expectation of this inequality we get

gif

the reason is

gif

which gives the Markov’s inequality for a>0 as

gif

Chebyshev’s inequality

 For the finite mean and variance of random variable X the Chebyshev’s inequality for k>0 is

gif

where sigma and mu represents the variance and mean of random variable, to prove this we use the Markov’s inequality as the non negative random variable

gif

for the value of a as constant square, hence

gif

this equation is equivalent to

gif

as clearly

gif

Examples of Markov’s and Chebyshev’s inequalities :

  1. If the production of specific item is taken as random variable for the week with mean 50 , find the probability of production exceeding 75 in a week and what would be the probability if the production of a week is between 40 and 60 provided the variance for that week is 25?

Solution: Consider the random variable X for the production of the item for a week then to find the probability of production exceeding 75 we will use Markov’s inequality as

gif

Now the probability for the production in between 40 to 60 with variance 25 we will use Chebyshev’s inequality as

gif

so

gif

this shows the probability for the week if the production is between 40 to 60 is 3/4.

2. Show that the chebyshev’s inequality which provides upper bound to the probability is not particularly nearer to the actual value of the probability.

Solution:

Consider the random variable X is uniformly distributed with mean 5 and variance 25/3 over the interval (0,1) then by the chebyshev’s inequality we can write

gif.latex?P%28%7CX

but the actual probability will be

gif.latex?P%28%7CX 5%7C%26gt%3B4%5C%7D%3D0

which is far from the actual probability likewise if we take the random variable X as normally distributed with mean and variance then Chebyshev’s inequality will be

gif

but the actual probability is

gif.latex?P%28%7CX %5Cmu%7C%26gt%3B2%20%5Csigma%5C%7D%3DP%5Cleft%5C%7B%5Cleft%7C%5Cfrac%7BX %5Cmu%7D%7B%5Csigma%7D%5Cright%7C%26gt%3B2%5Cright%5C%7D%3D2%5B1

Weak Law of Large Numbers

The weak law for the sequence of random variables will be followed by the result that Chebyshev’s inequality can be used as the tool for proofs for example to prove

gif

if the variance is zero that is the only random variables having variances equal to 0 are those which are constant with probability 1 so by Chebyshev’s inequality for n greater than or equal to 1

gif

as

gif

by the continuity of the probability

gif

which proves the result.

to prove this we assume that variance is also finite for each random variable in the sequence so the expectation and variance

gif

now from the Chebyshev’s inequality the upper bound of the probability as

gif

which for n tending to infinity will be

gif

Central Limit theorem

The central limit theorem is one of the important result in probability theory as it gives the distribution to the sum of large numbers which is approximately normal distribution in addition to the method for finding the approximate probabilities for sums of independent random variables central limit theorem also shows the empirical frequencies of so many natural populations exhibit bell-shaped means normal curves, Before giving the detail explanation of this theorem we use the result

“If the sequence of random variables Z1,Z2,…. have the distribution function and moment generating function as FZn and Mzn then

gif

Central Limit theorem: For the sequence of identically distributed and independent random variables X1,X2,……. each of which having the mean μ and variance σ2 then the distribution of the sum

gif

tends to the standard normal as n tends to infinity for a to be real values

Proof: To prove the result consider the mean as zero and variance as one i.e μ=0 & σ2=1 and the moment generating function for Xi exists and finite valued so the moment generating function for the random variable Xi/√n will be

gif

hene the moment generating function for the sum ΣXi/√n will be

gif

Now let us take L(t)=logM(t)

so

gif

to show the proof we first show

by showing its equivalent form

%202

since

hence this shows the result for the mean zero and variance 1, and this same result follows for the general case also by taking

%20%5Csigma

and for each a we have

gif

Example of the Central Limit theorem

To calculate the distance in light year of a star from the lab of an astronomer, he is using some measuring techniques but because of change in atmosphere each time the distance measured is not exact but with some error so to find the exact distance he plans to observe continuously in a sequence and the average of these distances as the estimated distance, If he consider the values of measurement identically distributed and independent random variable with mean d and variance 4 light year, find the number of measurement to do to obtain the 0.5 error in the estimated and actual value?

Solution: Let us consider the measurements as the independent random variables in sequence X1,X2,…….Xn so by the Central Limit theorem we can write

gif

which is the approximation into standard normal distribution so the probability will be

CodeCogsEqn 76

so to get the accuracy of the measurement at 95 percent the astronomer should measure n* distances where

gif.latex?2%20%5CPhi%5Cleft%28%5Cfrac%7B%5Csqrt%7Bn%5E%7B*%7D%7D%7D%7B4%7D%5Cright%29

so from the normal distribution table we can write it as

which says the measurement should be done for 62 number of times, this also can be observed with the help of Chebyshev’s inequality by taking

gif

so the inequality results in

gif.latex?P%5Cleft%5C%7B%5Cleft%7C%5Csum %7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7BX %7Bi%7D%7D%7Bn%7D d%5Cright%7C%26gt%3B0.5%5Cright%5C%7D%20%5Cleq%20%5Cfrac%7B4%7D%7Bn%280

hence for n=16/0.05=320 which gives certainity that there will be only o.5 percent error in the measurement of the distance of the star from the lab of observations.

2. The number of admitted students in engineering course is Poisson distributed with mean 100, it was decided that if the admitted students are 120 or more the teaching will be in two sections otherwise in one section only, what will be the probability that there will be two sections for the course?

Solution: By following the Poisson distribution the exact solution will be

gif

which is obviously not give the particular numerical value, If we consider the random variable X as the students admitted then by the central limit theorem

which can be

gif.latex?%5Cbegin%7Barray%7D%7Bl%7D%20%3DP%5Cleft%5C%7B%5Cfrac%7BX 100%7D%7B%5Csqrt%7B100%7D%7D%20%5Cgeq%20%5Cfrac%7B119.5 100%7D%7B%5Csqrt%7B100%7D%7D%5Cright%5C%7D%20%5C%5C%20%5Capprox%201

which is the numerical value.

3. Calculate the probability that the sum on ten die when rolled is between 30 and 40 including 30 and 40?

Solution: Here considering the die as Xi for ten values of i. the mean and variance will be

gif

thus following the central limit theorem we can write

gif.latex?%5Cbegin%7Baligned%7D%20P%5B29.5%20%5Cleq%20X%20%5Cleq%2040.5%5C%7D%20%26amp%3B%3DP%5Cleft%5C%7B%5Cfrac%7B29.5 35%7D%7B%5Csqrt%7B%5Cfrac%7B350%7D%7B12%7D%7D%7D%20%5Cleq%20%5Cfrac%7BX 35%7D%7B%5Csqrt%7B%5Cfrac%7B350%7D%7B12%7D%7D%7D%20%5Cleq%20%5Cfrac%7B40.5 35%7D%7B%5Csqrt%7B%5Cfrac%7B350%7D%7B12%7D%7D%7D%5Cright%5C%7D%20%5C%5C%20%26amp%3B%20%5Capprox%202%20%5CPhi%281.0184%29

which is the required probability.

4. For the uniformly distributed independent random variables Xi on the interval (0,1) what will be the approximation of the probability

gif

Solution: From the Unifrom distribution we know that the mean and variance will be

gif

Now using the central limit theorem we can

gif.latex?%5Cbegin%7Baligned%7D%20P%5Cleft%5C%7B%5Csum %7B1%7D%5E%7B10%7D%20X %7Bi%7D%26gt%3B6%5Cright%5C%7D%20%26amp%3B%3DP%5Cleft%5C%7B%5Cfrac%7B%5Csum %7B1%7D%5E%7B10%7D%20X %7Bi%7D 5%7D%7B%5Csqrt%7B10%5Cleft%28%5Cfrac%7B1%7D%7B12%7D%5Cright%29%7D%7D%26gt%3B%5Cfrac%7B6 5%7D%7B%5Csqrt%7B10%5Cleft%28%5Cfrac%7B1%7D%7B12%7D%5Cright%29%7D%7D%5Cright%5C%7D%20%5C%5C%20%26amp%3B%20%5Capprox%201

thus the summation of the random variable will be 14 percent.

5. Find the probability for the evaluator of the exam to give grades will be 25 exams in starting 450 min if there are 50 exams whose grading time is independent with mean 20 min and standard deviation 4 min.

Solution: Consider the time require to grade the exam by the random variable Xi so the random variable X will be

gif

since this task for 25 exam is withing 450 min so

gif
gif
gif

here using the central limit theorem

gif.latex?%5Cbegin%7Baligned%7D%20P%5BX%20%5Cleq%20450%5C%7D%20%26amp%3B%3DP%5Cleft%28%5Cfrac%7BX 500%7D%7B%5Csqrt%7B400%7D%7D%20%5Cleq%20%5Cfrac%7B450 500%7D%7B%5Csqrt%7B400%7D%7D%5Cright%29%20%5C%5C%20%26amp%3B%20%5Capprox%20P%28Z%20%5Cleq 2.5%5C%7D%20%5C%5C%20%26amp%3B%3DP%28Z%20%5Cgeq%202.5%5C%7D%20%5C%5C%20%26amp%3B%3D1 %5CPhi%282.5%29%3D0

which is the required probability.

Central Limit theorem for independent random variables

For the sequence which is not identically distributed but having independent random variables X1,X2,……. each of which having the mean μ and variance σ2 provided it satisfies

  1. each Xi is uniformly bounded
  2. sum of the variances is infinite, then
gif

Strong Law of Large Numbers

Strong Law of Large numbers is very crucial concept of the probability theory which says that the average of sequence of commonly distributed random variable with probability one will converge to the mean of that same distribution

Statement: For the sequence of identically distributed and independent random variables X1,X2,……. each of which having the finite mean with probability one then

gif

Proof: To prove this consider the mean of each of random variable is zero, and the series

gif

now for this consider power of this as

after taking the expansion of the right hand side terms we have the terms of the form

gif

since these are independents so the mean of these will be

gif

with the help of combination of the pair the expansion of the series now will be

gif

since

gif

so

gif

we get

gif

this suggest the inequality

gif

hence

gif

By the convergence of the series since the probability of each random variable is one so

gif

since

gif

if the mean of each random variable is not equal to zero then with deviation and probability one we can write it as

gif

or

gif

which is required result.

One Sided Chebyshev Inequality

The one sided Chebysheve inequality for the random variable X with mean zero and finite variance if a>0 is

Chebyshev's Inequality
chebyshev inequality

to prove this consider for b>0 let the random variable X as

gif

which gives

gif

so using the Markov’s inequality

Chebyshev's Inequality
one sided chebyshev

which gives the required inequality. for the mean and variance we can write it as

gif

This further can be written as

gif

Example:

Find the upper bound of the probability that the production of the company which is distributed randomly will at least 120, if the production of this certain company is having mean 100 and variance 400.

Solution:

Using the one sided chebyshev inequaility

gif

so this gives the probability of the production within a week atleast 120 is 1/2, now the bound for this probability will be obtained by using Markov’s inequality

which shows the upper bound for the probability.

Example:

Hundred pairs are taken from two hundred persons having hundred men and hundred women find the upper bound of the probability that at most thirty pair will consist a men and a women.

Solution:

Let the random variable Xi as

gif

so the pair can be expressed as

gif

Since every man can equally likely to be pair with remaining people in which hundred are women so the mean

gif

in the same way if i and j are not equal then

gif

as

%20197

hence we have

gif
gif
gif
gif
gif

using the chebyshev inequality

gif.latex?P%5C%7BX%20%5Cleq%2030%5C%7D%20%5Cleq%20P%5C%7B%7CX

which tells that the possibility of pairing 30 men with women is less than six, thus we can improve the bound by using one sided chebyshev inequality

gif.latex?%5Cbegin%7Baligned%7D%20P%5BX%20%5Cleq%2030%5C%7D%20%26amp%3B%3DP%5BX%20%5Cleq%2050.25

Chernoff Bound

If the moment generating function already known then

gif

as

gif

in the same way we can write for t<0 as

gif

Thus the Chernoff bound can be define as

gif

this inequality stands for all the values of t either positive or negative.

Chernoff bounds for the standard normal random variable

The Chernoff bounds for the standard normal random variable whose moment generating function

%202%7D

is

so minimizing this inequality and right hand side power terms gives for a>0

%202%7D

and for a<0 it is

%202%7D

Chernoff bounds for the Poisson random variable

The Chernoff bounds for the Poisson random variable whose moment generating function

gif

is

gif

so minimizing this inequality and right hand side power terms gives for a>0

%20%5Clambda 1%29%7D%5Cleft%28%5Cfrac%7B%5Clambda%7D%7Bi%7D%5Cright%29

and it would be

gif

Example on Chernoff Bounds

In a game if a player is equally likely to either win or lose the game independent of any past score, find the chernoff bound for the probability

Solution: Let Xi denote the winning of the player then the probability will be

gif

for the sequence of n plays let

gif

so the moment generating function will be

gif

here using the expansions of exponential terms

CodeCogsEqn 77

so we have

%202%7D

now applying the property of moment generating function

%202%7D%20%5Cend%7Baligned%7D

This gives the inequality

%202%20n%7D%20%5Cquad%20a%26gt%3B0

hence

Conclusion:

The inequalities and limit theorem for the large numbers were discussed and the justifiable examples for the bounds of the probabilities were also taken to get the glimpse of the idea, Also the the help of normal, poisson random variable and moment generating function is taken to demonstrate the concept easily, if you require further reading go through below books or for more Article on Probability, please follow our Mathematics pages.

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH

Point Sections Or Ratio Formulae: 41 Critical Solutions

by

Basic Examples on the Formulae “Point sections or Ratio”

Case-I

Problems 21:  Find the coordinates of the point P(x, y) which internally divides the line segment joining the two points (1,1) and (4,1) in the ratio 1:2.

Solution:   We already know,

If a point P(x, y) divides the line segment AB internally in the ratio m:n,where coordinates of A and B are (x1,y1) and (x2,y2) respectively. Then Coordinates of P are 

gif

and

gif

(See formulae chart)

Using this formula we can say , (x1,y1) ≌(1,1) i.e.   x1=1, y1=1   ;

(x2,y2)≌(4,1)   i.e.   x2=4, y2=1   

and

m:n  ≌ 1:2     i.e   m=1,n=2

Screenshot 4
Graphical Representation

Therefore,       

x  =

gif

( putting values of m & n in   

gif

Or, x  =1*4+2*1/3 ( putting values of x1 &  x2 too )

Or, x  = 4+2/3

Or, x  = 6*3

 Or, x = 2

Similarly we get,  

y =

gif

( putting values of m & n in     y  =

gif

Or, y =(1*1+2*1)/3 ( putting values of y1 &  y2 too )

Or, y = 1*1+2/3

Or, y  =  3/3

Or, y = 1

 Therefore, x=2 and y=1 are the coordinates of the point P i.e. (2,1).   (Ans)

More answered problems are given below for further practice using the procedure described in above problem 21:-

Problem 22: Find the coordinates of the point  which internally divides the line segment joining the two points (0,5) and (0,0) in the ratio 2:3.

                     Ans. (0,2)

Problem 23: Find the point which internally divides the line segment joining the points (1,1) and (4,1) in the ratio 2:1.

Ans. (3,1)

Problem 24: Find the point which lies on the line segment joining the two points (3,5,) and (3,-5,)  dividing it in the ratio 1:1

Ans. (3,0)

Problem 25: Find the coordinates of the point which internally divides the line segment joining the two points (-4,1) and (4,1) in the ratio 3:5

Ans. (-1,1)

Problem 26: Find the point which internally divides the line segment joining the two points (-10,2) and (10,2) in the ratio 1.5 : 2.5.

_____________________________

Case-II

Problems 27:   Find the coordinates of the point Q(x,y) which externally divides the line segment joining the two points (2,1) and (6,1) in the ratio 3:1.

Solution:  We already know,

If a point Q(x,y) divides the line segment AB externally in the ratio m:n,where coordinates of A and B are (x1,y1) and (x2,y2) respectively,then the coordinates of the point P are 

gif

and

gif

(See formulae chart)

Using this formula we can say ,  (x1,y1) ≌(2,1)  i.e.  x1=2, y1=1   ;

                                                    (x2,y2)≌(6,1)  i.e.   x2=6, y2=1    and   

                                                    m:n  ≌ 3:1    i.e.    m=3,n=1   

Point sections
Graphical Representation

Therefore, 

x =

gif

( putting values of m & n in     x  =

gif

Or, x =(3*6)-(1*2)/2 ( putting values of x1 &  x2 too )

Or, x = 18-2/2

Or, x  =16/2

Or, x = 8

Similarly we get,  

y =

gif

( putting values of m & n in     y  =

gif

Or, y  =

gif

( putting values of y1 &  y2 too )

Or, y = 3-1/2

Or, y  =  2/2

Or, y = 1

 Therefore, x=8 and y=1 are the coordinates of the point Q i.e. (8,1).   (Ans)

More answered problems are given below for further practice using the procedure described in above problem 27:-

Problem 28: Find the point which externally divides the line segment joining the two points (2,2) and (4,2) in the ratio 3 : 1.

Ans. (5,2)

Problem 29: Find the point which externally divides the line segment joining the two points (0,2) and (0,5) in the ratio 5:2.

Ans. (0,7)

Problem 30: Find the point which lies on the extended part of the line segment joining the two points (-3,-2) and (3,-2) in the ratio 2 : 1.

Ans. (9,-2)

________________________________

Case-III

Problems 31:  Find the coordinates of the midpoint  of the line segment joining the two points (-1,2) and (1,2).

Solution:   We already know,

If a point R(x,y) be the midpoint of the line segment  joining A(x1,y1) and B(x2,y2) .Then coordinates of R are

gif

and

gif

(See formulae chart)

Case-III is the form of case-I while m=1 and n=1

Using this formula we can say ,  (x1,y1) ≌(-1,2)  i.e.  x1=-1, y1=2   and

                                                    (x2,y2)≌(1,2)  i.e.   x2=1, y2=2

Screenshot 11
Graphical Representation

Therefore,

x =

gif

( putting values of x1 &  x2  in x =

gif

Or, x  = 0/2

Or, x = 0

Similarly we get, 

y =2+2/2 ( putting values of y1 &  y2  in y =

gif

Or, y = 4/2

Or, y = 2

Therefore, x=0 and y=2 are the coordinates of the midpoint R i.e. (0,2).   (Ans)

More answered problems are given below for further practice using the procedure described in above problem 31:-

Problem 32: Find the coordinates of the midpoint of the line joining the two points (-1,-3) and (1,-4).

Ans. (0,3.5)

Problem 33: Find the coordinates of the midpoint  which divides the line segment joining the two points (-5,-7) and (5,7).

Ans. (0,0)

Problem 34: Find the coordinates of the midpoint  which divides the line segment joining the two points (10,-5) and (-7,2).

Ans. (1.5, -1.5)

Problem 35: Find the coordinates of the midpoint  which divides the line segment joining the two points (3,√2) and (1,3√2).

Ans. (2,2√2)

Problem 36: Find the coordinates of the midpoint  which divides the line segment joining the two points (2+3i,5) and (2-3i,-5).

Ans. (2,0)

Note: How to check if a point divides a line (length=d units) internally or externally by the ratio m:n

If ( m×d)/(m+n)  +   ( n×d)/(m+n)  = d , then internally dividing and

If ( m×d)/(m+n)  –   ( n×d)/(m+n)  = d , then externally dividing

____________________________________________________________________________

Basic Examples on the Formulae “Area of a Triangle”

Case-I 

Problems 37: What is the area of the triangle with two vertices A(1,2) and B(5,3) and height with respect to AB be 3 units in the coordinate plane ?

 Solution:   We already know,

If “h” be the height and “b” be the base of Triangle, then  Area of the Triangle is  = ½ ×  b ×  h

(See formulae chart)

image?w=366&h=269&rev=57&ac=1&parent=1Ug0lE5AOAhO4i0HE5fVqVUKTEbR0on8yfNNyWgAF Po
Graphical Representation

Using this formula we can say , 

 h = 3 units and b = √

(x<sub>2</sub>-x<sub>1</sub>)<sup>2</sup>+(y<sub>2</sub>-y<sub>1</sub>)<sup>2 </sup>

i.e  √

(5-1)<sup>2</sup>+(3-2)<sup>2 </sup>

                    Or, b = √

(4)<sup>2</sup>+(1)<sup>2 </sup>

                    Or, b = √

(16+1<sup> </sup>

                    Or,  b = √ 17  units

Therefore, the required area of the triangle is   = ½ ×  b ×  h    i.e

= ½ × (√ 17 )  ×  3 units

= 3⁄2 × (√ 17 )   units   (Ans.)

______________________________________________________________________________________

Case-II

Problems 38:What is the area of the triangle with vertices A(1,2), B(5,3) and C(3,5) in the coordinate plane ?

 Solution:   We already know,

If  A(x1,y1) , B(x2,y2) and C(x3,y3) be the vertices of a Triangle,

Area of the Triangle is  =|½

</strong><strong>x</strong><strong><sub>1</sub></strong><strong> (y</strong><strong><sub>2</sub></strong><strong>-</strong><strong><sub> </sub></strong><strong> y</strong><strong><sub>3</sub></strong><strong>) + x</strong><strong><sub>2</sub></strong><strong> (y</strong><strong><sub>3</sub></strong><strong>-</strong><strong><sub> </sub></strong><strong> y</strong><strong><sub>2</sub></strong><strong>) + x</strong><strong><sub>3</sub></strong><strong> (y</strong><strong><sub>2</sub></strong><strong>- y</strong><strong><sub>1</sub></strong><strong>)</strong><strong>

|

(See formulae chart)

Using this formula we have , 

                                              (x1,y1) ≌(1,2) i.e.   x1=1, y1=2   ;

                                              (x2,y2) ≌(5,3)  i.e.   x2=5, y2=3 and

                                              (x3,y3) ≌(3,5)  i.e.    x3=3, y3=5

image?w=364&h=194&rev=207&ac=1&parent=1Ug0lE5AOAhO4i0HE5fVqVUKTEbR0on8yfNNyWgAF Po
Graphical Representation

Therefore, the area of the triangle is = |½

x<sub>1</sub> (y<sub>2</sub>-<sub> </sub> y<sub>3</sub>) + x<sub>2</sub> (y<sub>3</sub>-<sub> </sub> y<sub>1</sub>) + x<sub>3</sub> (y<sub>1</sub>-y<sub>2</sub>)

| i.e 

= |½

1 (3-5) + 5 (5-3) + 3 (3-2)

|  sq.units 

= |½

1x (-2) + (5×2) + (3×1)

|    sq.units

= |½

-2 + 10 + 3

|    sq.units

= |½ x 11|     sq.units

= 11⁄2     sq.units

= 5.5      sq.units         (Ans.)

More answered problems are given below for further practice using the procedure described in above problems :-

Problem 39: Find the area of the triangle whose vertices are (1,1), (-1,2) and (3,2).

Ans. 2 sq.units

Problem 40: Find the area of the triangle whose vertices are (3,0), (0,6) and (6,9).

Ans. 22.5 sq.units

Problem 41: Find the area of the triangle whose vertices are (-1,-2), (0,4) and (1,-3).

Ans. 6.5 sq.units

Problem 42: Find the area of the triangle whose vertices are (-5,0,), (0,5) and (0,-5).                                 Ans. 25 sq.units

 _______________________________________________________________________________________

For more post on Mathematics, please follow our Mathematics page.

Problems On Probability & Its Axioms

by
lagrida latex editor 33

Probability is a fundamental concept in mathematics that allows us to quantify uncertainty and make predictions about the likelihood of events occurring. It plays a crucial role in various fields, including statistics, economics, physics, and computer science. In this section, we will explore the definition of probability and its importance in mathematics, as well as the axioms that form the foundation of probability theory.

Definition of Probability and Its Importance in Math

Probability can be defined as a measure of the likelihood of an event occurring. It is represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The concept of probability is essential in mathematics because it helps us analyze and understand uncertain situations.

In real life, we encounter probabilistic situations every day. For example, when flipping a fair coin, we know that the probability of it landing on heads is 0.5. Similarly, when rolling a fair six-sided die, the probability of rolling a specific number, say 3, is 1/6. By understanding and applying probability, we can make informed decisions and assess risks in various scenarios.

Probability theory provides a systematic framework for studying and analyzing uncertain events. It allows us to mathematically model and analyze random phenomena, such as coin flips, dice rolls, and card games. By using probability theory, we can calculate the likelihood of different outcomes, estimate the expected value of random variables, and make predictions based on available data.

Axioms of Probability Theory

To ensure a consistent and coherent approach to probability, mathematicians have established a set of axioms that form the foundation of probability theory. These axioms provide a rigorous framework for defining and manipulating probabilities. Let’s take a closer look at the three axioms of probability:

  1. Non-negativity: The probability of any event is always a non-negative number. In other words, the probability of an event cannot be negative.

  2. Additivity: For any collection of mutually exclusive events (events that cannot occur simultaneously), the probability of the union of these events is equal to the sum of their individual probabilities. This axiom allows us to calculate the probability of complex events by considering the probabilities of their constituent parts.

  3. Normalization: The probability of the entire sample space (the set of all possible outcomes) is equal to 1. This axiom ensures that the total probability of all possible outcomes is always 1, providing a consistent framework for probability calculations.

By adhering to these axioms, we can ensure that our calculations and reasoning about probabilities are logically sound and consistent. These axioms, along with other probability concepts, such as conditional probability, independence, and Bayes’ theorem, form the building blocks of probability theory.

In the upcoming sections, we will delve deeper into probability theory, exploring various probability concepts, examples, exercises, and calculations. By understanding the axioms and principles of probability, we can develop a solid foundation for tackling more complex probability problems and applying probability in real-world scenarios.

Problems on Probability and Its Axioms

Example 1: Restaurant Menu Combinations

Imagine you’re at a restaurant with a diverse menu, offering a variety of appetizers, entrees, and desserts. Let’s say there are 5 appetizers, 10 entrees, and 3 desserts to choose from. How many different combinations of a meal can you create?

To solve this problem, we can use the fundamental principle of counting. The principle states that if there are m ways to do one thing and n ways to do another, then there are m * n ways to do both.

In this case, we can multiply the number of choices for each course: 5 appetizers * 10 entrees * 3 desserts = 150 different combinations of a meal.

Example 2: Probability of Item Purchases

Suppose you’re running an online store and you want to analyze the probability of customers purchasing certain items together. Let’s say you have 100 customers, and you track their purchase history. Out of these customers, 30 have bought item A, 40 have bought item B, and 20 have bought both items A and B. What is the probability that a randomly selected customer has bought either item A or item B?

To solve this problem, we can use the principle of inclusion-exclusion. This principle allows us to calculate the probability of the union of two events by subtracting the probability of their intersection.

First, we calculate the probability of buying item A or item B separately. The probability of buying item A is 30/100 = 0.3, and the probability of buying item B is 40/100 = 0.4.

Next, we calculate the probability of buying both item A and item B. This is given by the intersection of the two events, which is 20/100 = 0.2.

To find the probability of buying either item A or item B, we add the probabilities of buying each item and subtract the probability of buying both items: 0.3 + 0.4 – 0.2 = 0.5.

Therefore, the probability that a randomly selected customer has bought either item A or item B is 0.5.

Example 3: Probability of Card Occurrences

Let’s consider a standard deck of 52 playing cards. What is the probability of drawing a heart or a diamond from the deck?

To solve this problem, we need to determine the number of favorable outcomes (drawing a heart or a diamond) and the total number of possible outcomes (drawing any card from the deck).

There are 13 hearts and 13 diamonds in a deck, so the number of favorable outcomes is 13 + 13 = 26.

The total number of possible outcomes is 52 (since there are 52 cards in a deck).

Therefore, the probability of drawing a heart or a diamond is 26/52 = 0.5.

Example 4: Probability of Temperature Occurrences

Suppose you are interested in predicting the weather for the next day. You have observed that over the past year, the probability of a hot day is 0.3, the probability of a cold day is 0.2, and the probability of a rainy day is 0.4. What is the probability that tomorrow will be either hot or cold, but not rainy?

To solve this problem, we can use the probability addition rule. The rule states that the probability of the union of two mutually exclusive events is the sum of their individual probabilities.

In this case, the events “hot day” and “cold day” are mutually exclusive, meaning they cannot occur at the same time. Therefore, we can simply add their probabilities: 0.3 + 0.2 = 0.5.

Therefore, the probability that tomorrow will be either hot or cold, but not rainy, is 0.5.

Example 5: Probability of Card Denominations and Suits

Consider a standard deck of 52 playing cards. What is the probability of drawing a card that is either a king or a spade?

To solve this problem, we need to determine the number of favorable outcomes (drawing a king or a spade) and the total number of possible outcomes (drawing any card from the deck).

There are 4 kings and 13 spades in a deck, so the number of favorable outcomes is 4 + 13 = 17.

The total number of possible outcomes is 52 (since there are 52 cards in a deck).

Therefore, the probability of drawing a card that is either a king or a spade is 17/52 ≈ 0.327.

Example 6: Probability of Pen Colors

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Suppose you have a bag containing 5 red pens, 3 blue pens, and 2 green pens. What is the probability of randomly selecting a red or blue pen from the bag?

To solve this problem, we need to determine the number of favorable outcomes (selecting a red or blue pen) and the total number of possible outcomes (selecting any pen from the bag).

There are 5 red pens and 3 blue pens in the bag, so the number of favorable outcomes is 5 + 3 = 8.

The total number of possible outcomes is 5 + 3 + 2 = 10 (since there are 5 red pens, 3 blue pens, and 2 green pens in the bag).

Therefore, the probability of randomly selecting a red or blue pen from the bag is 8/10 = 0.8.

Example 7: Probability of Committee Formation

Suppose there are 10 people, and you need to form a committee of 3 people. What is the probability that you select 2 men and 1 woman for the committee?

To solve this problem, we need to determine the number of favorable outcomes (selecting 2 men and 1 woman) and the total number of possible outcomes (selecting any 3 people from the group of 10).

First, we calculate the number of ways to select 2 men from a group of 5 men: C(5, 2) = 10.

Next, we calculate the number of ways to select 1 woman from a group of 5 women: C(5, 1) = 5.

To find the total number of favorable outcomes, we multiply the number of ways to select 2 men by the number of ways to select 1 woman: 10 * 5 = 50.

The total number of possible outcomes is the number of ways to select any 3 people from a group of 10: C(10, 3) = 120.

Therefore, the probability of selecting 2 men and 1 woman for the committee is 50/120 ≈ 0.417.

Example 8: Probability of Suit Occurrences in a Card Hand

Consider a standard deck of 52 playing cards. What is the probability of drawing a hand of 5 cards that contains at least one card of each suit (hearts, diamonds, clubs, and spades)?

To solve this problem, we need to determine the number of favorable outcomes (drawing a hand with at least one card of each suit) and the total number of possible outcomes (drawing any hand of 5 cards from the deck).

First, we calculate the number of ways to select one card from each suit: 13 * 13 * 13 * 13 = 285,316.

Next, we calculate the total number of possible outcomes, which is the number of ways to draw any 5 cards from a deck of 52: C(52, 5) = 2,598,960.

Therefore, the probability of drawing a hand of 5 cards that contains at least one card of each suit is 285,316/2,598,960 ≈ 0.11.

Example 9: Probability of choosing the same letter from two words

When it comes to probability, we often encounter interesting problems that challenge our understanding of the subject. Let’s consider an example that involves choosing the same letter from two words.

Suppose we have two words, “apple” and “banana.” We want to determine the probability of randomly selecting the same letter from both words. To solve this problem, we need to break it down into smaller steps.

First, let’s list all the letters in each word:

Word 1: “apple”
Word 2: “banana”

Now, we can calculate the probability of choosing the same letter by considering each letter individually. Let’s go through the process step by step:

  1. Selecting a letter from the first word:
  2. The word “apple” has five letters, namely ‘a’, ‘p’, ‘p’, ‘l’, and ‘e’.

  3. The probability of selecting any particular letter is 1 out of 5, as there are five letters in total.

  4. Selecting a letter from the second word:

  5. The word “banana” has six letters, namely ‘b’, ‘a’, ‘n’, ‘a’, ‘n’, and ‘a’.

  6. Similarly, the probability of selecting any particular letter is 1 out of 6.

  7. Calculating the probability of choosing the same letter:

  8. Since each letter has an equal chance of being selected from both words, we multiply the probabilities together.
  9. The probability of selecting the same letter is (1/5) * (1/6) = 1/30.

Therefore, the probability of choosing the same letter from the words “apple” and “banana” is 1/30.

What are the important properties of conditional expectation and how do they relate to problems on probability and its axioms?

The concept of conditional expectation is a fundamental concept in probability theory, and it has important properties that can help us solve problems related to probability and its axioms. To understand these properties and their relationship to probability problems, it is essential to delve into the Properties of conditional expectation explained. These properties provide insights into how conditional expectations behave and can be used to calculate expectations and probabilities in various scenarios. By understanding these properties, we can bridge the gap between the concept of probability and its axioms and the idea of conditional expectation, enabling us to tackle complex probability problems with confidence.

Frequently Asked Questions

1. What is the importance of probability in math?

Probability is important in math because it allows us to quantify uncertainty and make predictions based on available information. It provides a framework for analyzing and understanding random events and their likelihood of occurrence.

2. How would you define probability and its axioms?

Probability is a measure of the likelihood of an event occurring. It is defined using three axioms:

  1. The probability of any event is a non-negative number.
  2. The probability of the entire sample space is 1.
  3. The probability of the union of mutually exclusive events is equal to the sum of their individual probabilities.

3. What are the three axioms of probability?

The three axioms of probability are:

  1. Non-negativity: The probability of any event is a non-negative number.
  2. Normalization: The probability of the entire sample space is 1.
  3. Additivity: The probability of the union of mutually exclusive events is equal to the sum of their individual probabilities.

4. What are the axioms of expected utility theory?

The axioms of expected utility theory are a set of assumptions that describe how individuals make decisions under uncertainty. They include the axioms of completeness, transitivity, continuity, and independence.

5. What are the axioms of probability theory?

The axioms of probability theory are the fundamental principles that govern the behavior of probabilities. They include the axioms of non-negativity, normalization, and additivity.

6. Can you provide some solved problems on axioms of probability?

Certainly! Here is an example:

Problem: A fair six-sided die is rolled. What is the probability of rolling an even number?

Solution: Since the die is fair, it has six equally likely outcomes: {1, 2, 3, 4, 5, 6}. Out of these, three are even numbers: {2, 4, 6}. Therefore, the probability of rolling an even number is 3/6 = 1/2.

7. Where can I find probability problems and answers?

You can find probability problems and answers in various resources such as textbooks, online math websites, and educational platforms. Additionally, there are specific websites that provide probability problems and solutions, such as Math-Aids Answers.

8. Are there any probability examples available?

Yes, there are many probability examples available. Some common examples include flipping a coin, rolling dice, drawing cards from a deck, and selecting balls from an urn. These examples help illustrate how probability concepts can be applied in different scenarios.

9. What are some probability formulas and rules?

There are several probability formulas and rules that are commonly used, including:

  • Addition Rule: P(A or B) = P(A) + P(B) – P(A and B)
  • Multiplication Rule: P(A and B) = P(A) * P(B|A)
  • Complement Rule: P(A’) = 1 – P(A)
  • Conditional Probability: P(A|B) = P(A and B) / P(B)
  • Bayes’ Theorem: P(A|B) = P(B|A) * P(A) / P(B)

10. Can you suggest some probability exercises for practice?

Certainly! Here are a few probability exercises you can try:

  1. A bag contains 5 red balls and 3 blue balls. What is the probability of drawing a red ball?
  2. Two dice are rolled. What is the probability of getting a sum of 7?
  3. A deck of cards is shuffled and one card is drawn. What is the probability of drawing a heart?
  4. A jar contains 10 red marbles and 5 green marbles. If two marbles are drawn without replacement, what is the probability of getting two red marbles?
  5. A spinner is divided into 8 equal sections numbered 1 to 8. What is the probability of landing on an even number?

These exercises will help you practice applying probability concepts and calculations.