edited by
702 views
1 votes
1 votes
Suppose you have two coins $A$ and $B$ the probability of head in $A$ is $\dfrac{1}{4}$ and the probability of head in $B$ is $\dfrac{3}{4}$.

Now, suppose you have chosen a coin and tossed it two times.

The output was head and head. What is the probability that you chose the coin $B$.
edited by

3 Answers

Best answer
7 votes
7 votes

Here , Probability that it is coin B and output is HH after tossing coin 'B' 2 times , P(HH | B ) = (3/4)*(3/4) = 9/16

and     Probability that it is coin A and output is HH after tossing coin 'A' 2 times, P(HH | A ) = (1/4)*(1/4) = 1/16

now , Probability of choosing coin A = P(A) = 1/2

and    Probability of choosing coin B  = P(B) = 1/2

here choosing coins 'A' and 'B' are disjoint events...

Now , using Bayes' Theorem ,

P(B | HH) = P(HH | B)*P(B) / [P(HH | A)*P(A) + P(HH | B)*P(B)]

                = (9/16)*(1/2) / [(1/16)*(1/2) + (9/16)*(1/2)]

                = 9/10

So, Answer is 9/10

selected by
1 votes
1 votes

i think it is (1/10).

it is based on Reverse Probability (Bayes Theorem).



Sorry , i have given answer based on A  has to chose.

edited by
1 votes
1 votes
when coin B is tossed twice, P(HH)=$\frac{3}{4}*\frac{3}{4}$

when coin A is tossed twice, P(HH)=$\frac{1}{4}*\frac{1}{4}$

according to Bayes Theorem,

Probability of getting both the heads when coin B is chosen=$\frac{\frac{3}{4}*\frac{3}{4}}{\frac{1}{4}*\frac{1}{4}+\frac{3}{4}*\frac{3}{4}}$=$\frac{9}{10}$

Related questions

0 votes
0 votes
1 answer
3
Sayan Bose asked Apr 14, 2019
649 views