Question

Given $3$ identical boxes $1,2,$ & $3$. Each containing $2$ coins. In box $1$, both are gold coins & in box $2$ both are silver coins and in box 3 there is $1$ silver and $1$ gold coin. A person chooses a box at random and takes out the coin. If the coin is gold then what is the probability that it is coming from the first box?

Answer 1

Let $E_1,E_2,E_3$ be the event that boxes 1, 2 and 3 are chosen respectively.

Then,

$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

Let A be the event that the coin drawn is of gold.

Then, $P\left(A|E_1\right)=\frac{2}{2}=1$

$P\left(A|E_2\right)=0$

$P\left(A|E_3\right)=\frac{1}{2}$

Probability that other gold coin is drawn from Bag 1 = $P\left(E_1|A\right)$

By Baye's theorem,

$P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$

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

Hence, required probability is $\frac{2}{3}$.