event $B_2$ : Selecting box II having two silver coins
event $B_3$ : Selecting box III having one silver and one gold coin,
event G: Coin is gold.
$P \left( B _1\right)= P \left( B _2\right)= P \left( B _3\right)=\frac{1}{3}$
$\begin{aligned} & P \left( G / B _1\right)=1, P \left( G / B _2\right)=0, P \left( G / B _3\right)=\frac{1}{2} \\ & P ( G )= P _{(}\left( B _1\right) P \left( G / B _1\right)+ P \left( B _2\right) P \left( G / B _2\right)\end{aligned}$
$+ P \left( B _3\right) P \left( G / B _3\right)$
$=\frac{1}{3}\left[1+0+\frac{1}{2}\right]=\frac{1}{3}\left(\frac{3}{2}\right)=\frac{1}{2}$
To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from the box I.
∴ Required probability = P(B1/G)
By Bayes’ theorem,
$P\left(B_1 / G\right)=\frac{P\left(B_1\right) P\left(G / B_1\right)}{P(G)}$
$=\frac{\frac{1}{3}(1)}{\frac{1}{2}}=\frac{2}{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Weights | Heights | |
| Mean | 63.2kg | 63.2 inch |
| Standard deviation | 5.6kg | 11.5 inch |