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

Answer: (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,

$P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)+P\left(E_3\right)P\left(A|E_3\right)}$
$=\frac{\frac{1}{3}\cdot 1}{\frac{1}{3}\cdot 1+\frac{1}{3}\cdot 0+\frac{1}{3}\cdot \frac{1}{2}}$
$=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$.