Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

$\frac{2}{3}$

The correct answer is Option (3) → $\frac{2}{3}$ ##

Let $A, B$ and $C$ be the events that boxes I, II and III are chosen, respectively.

$∴\quad P(A) = P(B) = P(C) = \frac{1}{3}$

Let $E$ be the event that "the coin drawn is of gold".

Then $P(E|A) = P(\text{a gold coin from box I}) = \frac{2}{2} = 1$

$P(E|B) = P(\text{a gold coin from box II}) = 0$

$P(E|C) = P(\text{a gold coin from box III}) = \frac{1}{2}$

Now, by Bayes' theorem, the probability that gold coin is drawn from the box I

$P(A|E) = \frac{P(A)P(E|A)}{P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C)}$

$= \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times 0 + \frac{1}{3} \times \frac{1}{2}} = \frac{2}{3}$