Question

Box-I contains 2 gold coins, while another Box-II contains 1 gold and 1 silver coin. A person choose a box at random and takes out a coin. If the coin is of gold, what is the probability that the order coin in the box is also of gold?

Answer 1

Let $E_1$ and $E_2$ be the events that the boxes $I$ and $II$ are chosen respectively.

$⟹P\left(E_1\right)=\frac{1}{2}=P\left(E_2\right)$

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

$⟹P\left(A|E_1\right)=P\left(\text{a gold coin from box I}\right)=\frac{2}{2}=1$

$P\left(A|E_2\right)=P\left(\text{a gold coin from box II}\right)=\frac{1}{2}$

Probability that the other coin in the box is of gold $=$ The probability that the gold coin is drawn from box $\text{I}=P\left(E_1|A\right)$

By using Bayes' theorem,

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

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

$=\frac{2}{3}$

Hence, $P\left(A|E_1\right)=\frac{2}{3}$