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# ISI MStat PSB 2006 Problem 5 | Binomial Distribution

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 5 based on use of binomial distribution . Let's give it a try !!

## Problem- ISI MStat PSB 2006 Problem 5

Suppose $X$ is the number of heads in 10 tossses of a fair coin. Given $X=5,$ what is the probability that the first head occured in the third toss?

### Prerequisites

Basic Counting Principle

Conditional Probability

Binomial Distribution

## Solution :

As $X$ is the number of heads in 10 tossses of a fair coin so $X \sim binom(10, \frac{1}{2} )$

A be the event that first head occured in third toss

B be the event that X=5

We have to find that $P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{ {7 \choose 4} {\frac{1}{2}}^{10} }{ {10 \choose 5} {\frac{1}{2}}^{10}}$

As , $P(A \cap B)$ = Probability that out of 5 heads occur at 10 tosses 1st head occur at 3rd throw

=Probability that first two tails $\times$ probability that 3rd one is head $\times$ probability that out of 7 toss 4 toss will give head

= ${\frac{1}{2}}^2 \times \frac{1}{2} \times {7 \choose 4} {\frac{1}{2}}^{7}$

Hence our required probability is $\frac{5}{36}$

## Food For Thought

Under the same condition find the probability that X= 3 given 1st head obtained from 2nd throw .

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This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 5 based on use of binomial distribution . Let's give it a try !!

## Problem- ISI MStat PSB 2006 Problem 5

Suppose $X$ is the number of heads in 10 tossses of a fair coin. Given $X=5,$ what is the probability that the first head occured in the third toss?

### Prerequisites

Basic Counting Principle

Conditional Probability

Binomial Distribution

## Solution :

As $X$ is the number of heads in 10 tossses of a fair coin so $X \sim binom(10, \frac{1}{2} )$

A be the event that first head occured in third toss

B be the event that X=5

We have to find that $P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{ {7 \choose 4} {\frac{1}{2}}^{10} }{ {10 \choose 5} {\frac{1}{2}}^{10}}$

As , $P(A \cap B)$ = Probability that out of 5 heads occur at 10 tosses 1st head occur at 3rd throw

=Probability that first two tails $\times$ probability that 3rd one is head $\times$ probability that out of 7 toss 4 toss will give head

= ${\frac{1}{2}}^2 \times \frac{1}{2} \times {7 \choose 4} {\frac{1}{2}}^{7}$

Hence our required probability is $\frac{5}{36}$

## Food For Thought

Under the same condition find the probability that X= 3 given 1st head obtained from 2nd throw .

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