Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc Data Science.  Learn More 

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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
rockethighlight