Suppose you have two coins which appear identical in your pocket. You know that, one is fair and one is 2 headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

$1/3$

The correct answer is Option (2) → $1/3$ ##

Let $E_1 = \text{Event that fair coin is drawn, we get}$

$ E_2 = \text{Event that 2 headed coin is drawn}$

$ E = \text{Event that tossed coin get a head}$

$∴P(E_1) = 1/2, P(E_2) = 1/2, P(E | E_1) = 1/2 \text{ and } P(E | E_2) = 1$

Now, using Baye's theorem,

$P(E_1 | E) = \frac{P(E_1) \cdot P(E | E_1)}{P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2)}$

$= \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1}$

$= \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$