Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The
number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is :
Correct Option: 1
Let $\mathrm{B}_{1}$ be the event where Box-I is selected. $\& \mathrm{~B}_{2} \rightarrow$ where box-II selected
$\mathrm{P}\left(\mathrm{B}_{1}\right)=\mathrm{P}\left(\mathrm{B}_{2}\right)=\frac{1}{2}$
Let $\mathrm{E}$ be the event where selected card is non prime.
For $\mathrm{B}_{1}$ : Prime numbers :
$\{2,3,5,7,11,13,17,19,23,29\}$
For $\mathrm{B}_{2}$ : Prime numbers :
$\{31,37,41,43,47\}$
$\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{B}_{1}\right) \times \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{B}_{1}}\right)+\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{B}_{2}}\right)$
$=\frac{1}{2} \times \frac{20}{30}+\frac{1}{2} \times \frac{15}{20}$
Required probability :
$P\left(\frac{B_{1}}{E}\right)=\frac{\frac{1}{2} \times \frac{20}{30}}{\frac{1}{2} \times \frac{20}{30}+\frac{1}{2} \times \frac{15}{20}}=\frac{\frac{2}{3}}{\frac{2}{3}+\frac{3}{4}}=\frac{8}{17}$