Question

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 box $III$, there is one gold coin and one silver coin. $A$ person chooses a box at random and takes out a coin. If the coin is of gold, then what is the probability that the other coin in the box is also of gold?

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{2}{3}$

Answer 1

The Correct Option is D

Let $E_1$, $E_2$, $E_3$ and $A$ be the events defined as follows : $E_1 =$ box $I$ is chosen, $E_2 =$ box $II$ is chosen, $E_3 =$ box $III$ is chosen and $A =$ a gold coin has been taken out Then $P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$ $P(A|E_1) = P$( drawing a gold coin from box $I$) $= \frac{2}{2}= 1$ $P(A|E_2) = P$(drawing a gold coin from box $II$) $= \frac{0}{2} = 0$ $P(A |E_3) = P$(drawing a gold coin from box $III$) = $\frac{1}{2}$ We want to find the probability that the other coin in the chosen box is gold i.e., the probability that gold coin is drawn from box $I$ By Bayes' theorem, $P\left(E_{1}|A\right) = \frac{P\left(E_{1}\right)P\left(A |E_{1}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)+P\left(E_{3}\right)P\left(A|E_{3}\right)}$ $= \frac{\frac{1}{3}\cdot1}{\frac{1}{3}\cdot1+\frac{1}{3}\cdot0+\frac{1}{3}\cdot\frac{1}{2}} $ $= \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.