Question

Each of three identical jewellery boxes has two drawers. In each drawer of the first box, there is a gold watch. In each drawer of the second box, there is a silver watch. In one drawer of the third box, there is a gold watch while in the other, there is a silver watch. If we select a box at random, open one of the drawers and find it to contain a silver watch, then the probability that the other drawer has the gold watch in it, is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{1}{3}$



$A:$ One box and one of its drawers randomly selected and a silver watch found in it
$B_1:$ It is the box $C$
$B_2:$ It is the box $B$
$B_3:$ It is the box $A$

$P\left(B_1\right)=P\left(B_2\right)=P\left(B_3\right)=\frac{1}{3}$
$P\left(A|B_1\right)=\frac{1}{2};P\left(A|B_2\right)=1;P\left(A|B_3\right)=0$
$P\left(B_1|A\right)=\frac{P\left(A|B_1\right)\times P\left(B_1\right)}{P\left(A|B_1\right)\times P\left(B_1\right)+P\left(A|B_2\right)\times P\left(B_2\right)+P\left(A|B_3\right)\times P\left(B_3\right)}$
$\Rightarrow P\left(B_1|A\right)=\frac{\frac{1}{2}}{\frac{1}{2}+1}=\frac{1}{2}\cdot \frac{2}{3}=\frac{1}{3}$