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Total Probability and Bayes Theorem

Total Probability (1)

It is an application of Conditional Probability and Dependent Event.

We have already discussed conditional probability and Independent events.

Dependent Events :

Events that are depending on each other, here the sequence in which the events are happening is important. Example:

let us say a bag has 4 green balls and 7 red balls and we are pulling two balls on after other. 

What is the probability that both balls are Red?

Case 1: Replacement is allowed :
 

Solution:

for the first ball: P(red1) = 7 /11

for the second ball :

Since we are putting the first ball then again the total number of ball = 11

So P(red2) = 7/11

So P(red1 and red2) = (7/11) *(7/11) = 49/121

Case 2: No replacement: 

Here we are not putting the first ball back in the bag.

P(red1 ∩ red 2 )=  P(red1) * P(red2 / red1)

                              = (7/11) * (6/10)

                              = 42/110

This is the case of dependent events because when you are trying to pull the second ball it depends on the first event.

 

Total Probability example 1:-

let's say we have two bags and bag1 contains 4 red and 3 green and bag2 contains  5 red and 6 green balls

find out what is the probability of drawing a red ball? 

Solution:- Now drawing a red ball depends on two factors:

1. First, we should know from which bag that ball is drawn and what is the probability of choosing that bag?

2. Composition of that chosen bag?

 

Let's say the probability of choosing each bag is 1/2

 So the probability of drawing a red ball  P(red) given as:

P(red )= Bag 1 is selcted and the red ball is drawn OR Bagis selcted and red ball is drawn

           {  P( Bag 1 ). (Pred /Bag 1 ) }+  {P(Bag). P(Pred / Bag2)}

            = (1/2)(4/7) + (1/2)(5/11)

            = ( 4/14 ) + (5/22)

            

 

 

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Total Probability (2)

In a multiple choice question Exam, a student either knows the answer or he guesses. There are m options available in each question. If he guesses the answer, he is correct with 1/m probability.

Suppose p is the probability that he knows the answer and 1-p is the probability he doesn't know.

 

What is the probability that he has given answer correctly?

This is a case of Total Probability.

P(correct) = P(he knowns answer and answer is correct) + P(he guess the answer and answer is correct)

                      = p(1) + (1-p)*(1/m)

 

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Bayes' theorem Introduction

let's say we have two bags and each bag contains some red balls and some green balls. One ball is randomly chosen and we are asked to find the probability that the chosen ball is red.

 

It is very easy to answer this question. We need to apply total probability theorem and we will get our answer.

But what if they already mentioned that a red ball is chosen and we are asked to find the probability that the chosen ball is from the first bag.

 

let's understand this with an example:

1. let us say we have three bags :

Question 1: What is the probability of choosing a red ball?

Question 2: What is the probability that the chosen red ball is from bag A?

Please read these questions carefully.

Solution 1: In this, they are simply asking to find out how we are going to pick a red ball...

let's say the probability of selecting any bag is 1/3

So P(R) = P(BA and R) + P(Band R) ................ (1)

for Question 2:

They have already mentioned that a red ball is chosen already and we have to find out the probability that it is chosen from bag  'A'.

It can be written as P( A/R ).

P(A/R)= P(A ∩R)/ P(R) 

             = P(A ∩R) / { P(BA and R) + P(Band R)} ....... puting the value of P(R) from equation 1

 

The formal definition of Baye's Theorem:

             

Derivation Of Baye's Theorem:-

 Now we will solve some numerical based on Baye's Theorem.

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Baye's Theorem Example 1

Bag A contains 3 red balls and 4 black balls while Bag B contains 5 red balls and 6 black balls.

One ball is drawn and it is found to be red. Find the probability that it was drawn from bag 'B' .

 

Solution:

P(B/R) = ??

we know P(B/R) = P(B∩R) / P(R).....(1)

and P(R/B) = P(B∩R) / P(B)......(2)

by using equation 2 we can say tha

P(B/R) ={ P(B). P(R/B) } / P(R)

and we alreday know how to calculate P(R) . 

P(R)= P(B).(R/B) + P(A).P(R/B)

        ={(1/2).(5/11) +(1/2).(3/7) } 

now 

P(B/R)= { (1/2).(5/11) } /  {(1/2).(5/11) +(1/2).(3/7) } 

 

 

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Section 3.3 Example 4f Sheldon Ross on Baye's Theorem.

An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at some time within a fixed 1-year period with probability .4, whereas this probability decreases to .2 for a person who is not accident prone.

a) If we assume that 30 percent of the population is accident prone, what is the probability that a new policyholder will have an accident within a year of purchasing a policy?

b) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone?

a)

We shall obtain the desired probability by first conditioning upon whether or not the policyholder is accident prone. Let A1 denote the event that the policyholder will have an accident within a year of purchasing the policy, and let A denote the event that the policyholder is accident prone. Hence, the desired probability is given by:

 

\(P(A_1)\) = \(P(A_1 | A)P(A) + P(A_1 | A^c)P(A^c)\)
\(= (.4)(.3) + (.2)(.7) = .26\)

b) The desired probability is

\(P(A | A_1) = \frac{P(AA_1)}{P(A_1)} = \frac{P(A)P(A_1| A)}{P(A_1)} = \frac{(.3)(.4)}{.26} = \frac{6}{13}\)

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Baye's Theorem Example 2

There are three identical boxes A, B and C  each having two coins.

In box A, both are gold coins.

In box B, both are silver coins.

In box C, one is gold other is a sliver.

If a coin is picked up and found to be gold. Find the probability that the other coin in the box is gold.

 

Solution:

They are saying to find the probability that other coins is also gold.  this is possible only when box A is selected because it contains two gold coins. So this question can be also asked as "find the probability that gold coin is drawn from bag A".

We have to find out P(A/ G) =?

P(A/G) = P(A∩ G) / P(G) or

              ={ P(A). P(G/A) } / P(G)

Probabilty of choosing a bag is 1/3 beacuse there are 3 bags and all are identical

P(G)= P(A).P(G/A) + P(B).P(G/B) + P(C).P(G/C)

          = (1/3). (1) + (1/3)(0) + (1/3)(1/2)

         =   1/2 

P(A). P(G/A) = 1/3

P(A/ G) = (1/3) / (1/2)= 2/3

 

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