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Q.10 You are given three coins: one has heads on both faces, the second has tails on both faces, and the
third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it
comes up heads. The probability that the other face is tails is
(A) 1/4 (B) 1/3 (C) 1/2 (D) 2/3

In this question they have given answer (B) as 1/3. We can get B if we take

P(C1) = 1/3, P(C2) = 1/3,P(C3) = 1/3

Issue is that if we know that we have got heads , we can actually eliminate P(C3) , we can take it as 0.

Because it is given that we get Head when we toss it. (C3 can't generate head)

If we consider P(C1) = 1/2, P(C2) = 1/2 & P(C3) = 0 then answer I get is (D) 2/3 which is wrong as per GATE key. So please answer the question & Also let me know why should we consider C3 , if we know surely that coin is not C3.

I think we can't eliminate it from the selection. Because we were not knowing at time of the picking up coin that it need to have the heads. So any one out of three can be picked up.

We are given that head comes on one side. So, there are only two possibilities- the other side is head ($\text{HH}$ coin) or the other side is tail ($\text{HT}$ coin). Favorable case is only $\text{HT}$, and hence answer is $\frac{1}{2}$.

Well, the above explanation is wrong because it assumes uniform probability distribution for the two cases $\text{HH}$ and $\text{HT}$. This is true initially as all 3 coins have equal probability of being chosen. But we are given that "head" comes in one face. $\text{HH}$ coin has two head faces, and hence has twice the chance of being chosen over $\text{HT}$ as we got a head face. i.e., for the two coins $\text{HH}$ and $\text{HT}$, the respective probabilities are $\frac{2}{3}$ and $\frac{1}{3}$ respectively given that one side is a head. Now, our required probability is simply $P(\{\text{HT}\}) = \frac{1}{3}$.

We can also apply Bayes' theorem but I guess solving intuitively is more fun :)
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@Arjun, I am facing issue here, which topic I should learn for understanding that probability of HH is 2/3 ? Confused !
HH, HT are two coins. One has two head faces and other has 1. So, if a head face comes, probability of it being HH is 2 times the probability for other.
:( It was so simple :( How I missed it I wonder !

P(head) = 1(Doubly Headed coin) + 0 (Doubly tailed coin) + 1/2 (Normal coin)
=1/3