Question

Given $3$ identical boxes $I,II$ and $III$ each containing two coins. In box $I$ both are gold coins, in box $II$ both are silver coins and in box $III$ one is gold and $1$ is silver. A coin is drawn from one of the boxes. If the coin is of gold, what is the probability that the other coin is the box is also of gold.(write answer up-to 2 decimal place)

Answer 1

Let $B_1$: selecting box $1$ having two gold coins.

$B_2$: selecting box $2$ having two silver coins

$B_3$: selecting box $3$ having one gold & one silver.

G: The second coin is of Gold

We need to find the probability that the other coin in the box is also of gold, if the first coin is of gold

i.e., $P\left(B_1/G\right)$

$P\left(B_1/G\right)=\frac{P\left(B_1\right)\cdot P\left(G/B_1\right)}{P\left(B_1\right)\cdot P\left(G/B_1\right)+P\left(B_2\right)\cdot P\left(G/B_2\right)+P\left(B_3\right)P\left(G/B_3\right)}$

$P\left(B_1\right)=$Probability of selecting box $1$

$=1/3$

$P\left(G/B_1\right)=$ Probability that second coin is of gold in box $1=1$

$P\left(B_2\right)=$Probabiluty of selecting box $2$

$=1/3$

$P\left(G/B_2\right)=$Probability that second coin is of gold in box $2$

$=0$

$P(B{)}_3=$Probability of selecting box $3$

$=1/3$

$P\left(G/B_3\right)=$ Probability that second coin is of gold in box $3$

$=1/2$

$P\left(B_1/G\right)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+0\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{2}}$

$=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}}$

$P\left(B_1/G\right)=\frac{2}{3}$.