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 !!
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?
Basic Counting Principle
Conditional Probability
Binomial Distribution
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 \( \frac{5}{36} \)
Under the same condition find the probability that X= 3 given 1st head obtained from 2nd throw .
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 !!
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?
Basic Counting Principle
Conditional Probability
Binomial Distribution
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 \( \frac{5}{36} \)
Under the same condition find the probability that X= 3 given 1st head obtained from 2nd throw .
