Question

Given three identical boxes $I,II\text{and}III$, each containing two coins. In box $I$, both coins are gold coins, in box coins are gold coins, in box $II$, both are silver coins, in the box $III$, there is one gold and one silver coin. A person chooses a box at coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Answer 1

Given:
Let
$B_1:\text{be the event of choosing the box I}$.
$B_2:$ be the event of choosing the box $II$
$B_3:$ be the event of choosing the box $III$
and
$E:$ be the event that a gold coin is drawn

Since all the boxes are equally likely to be selected then,
Probability of selecting box -$I,II\text{and}III$
$P\left(B_1\right)=P\left(B_2\right)=P\left(B_3\right)=\frac{1}{3}$
Probablity of drawing a gold coin from box $I$

$P\left(E|B_1\right)=\frac{2}{2}=1$
Probablity of drawing agold coin from box $II$
$P\left(E|B_2\right)=\frac{0}{2}$
$P\left(E|B_2\right)$ = 0
Probability of drawing a gold coin from Box $III$
$P\left(E|B_3\right)=\frac{1}{2}$
By Bayes, theorem,
$P\left(B_1|E\right)$=
$\frac{P\left(B_1\right).P\left(E|B_1\right)}{P\left(B_1\right).P\left(E|B_1\right)+P\left(B_2\right).P\left(E|B_2\right)+P\left(B_2\right).P\left(E|B_3\right)}$ ...$\left(1\right)$

Substituting value of $P\left(B_1\right),P\left(B_2\right),P\left(B_3\right),P\left(E|B_1\right),P\left(E|B_2\right)\&P\left(E|B_3\right)$ in $\left(1\right)$,

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

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

$P\left(B_1|E\right)=\frac{1}{1+\frac{1}{2}}$

$P\left(B_1|E\right)=\frac{1}{\frac{3}{2}}$

$P\left(B_1|E\right)=\frac{2}{3}$

Therefore, probability that the other coin in the box is also of gold= $\frac{2}{3}$