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

Prerequisites

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 .

Similar Problems and Solutions

• 2019 PSB Problem 4
• 2019 PSB Problem 6
• Grasshopper Problem in Parity - Video

Related Program