CMI2018-A-6

Question

You are given two coins $A$ and $B$ that look identical. The probability that coin $A$ turns up heads is $\frac{1}{4}$, while the probability that coin $B$ turns up heads is $\frac{3}{4}.$ You choose one of the coins at random and toss it twice. If both the outcomes are heads, what is the probability that you chose coin $B?$

• A. $\frac{1}{16}$
• B. $\frac{1}{2}$
• C. $\frac{9}{16}$
• D. $\frac{9}{10}$

Comments

can we solve this problem by combination of Total Probability and bayes theorem

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9/10

Answer 1

$P(getting\ a\ head\ from \ coin \ A) = \frac{1}{4}$

$P(getting\ a\ head\ from \ coin \ B) = \frac{3}{4}$

Probability of choosing $A =P(A)= \frac{1}{2}$

Probability of choosing $B =P(B) = \frac{1}{2}$

If we choose $A$ the probability of getting head twice $=P(H/A)= \frac{1}{4} * \frac{1}{4} = \frac{1}{16}$

If we choose $B$ the probability of getting head twice $=P(H/B)= \frac{3}{4} * \frac{3}{4} = \frac{9}{16}$

$P(B/H) = ? $

So using Baye's theorem,

$P(B/H) = \frac{P(H/B) *P(B)}{ P(H/A) *P(A) + P(H/B) *P(B) } = \frac{\frac{9}{16} *\frac{1}{2}}{ \frac{1}{16} *\frac{1}{2} + \frac{9}{16} *\frac{1}{2} } = \frac{9}{10}$

So option $D$ is the correct answer.

Answer 2