Box I contains 30 cards numbered 1 to 30 and Box II

Question:

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 :

  1. $\frac{8}{17}$

  2. $\frac{2}{3}$

  3. $\frac{4}{17}$

  4. $\frac{2}{5}$


Correct Option: 1

Solution:

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 :

$P\left(\frac{B_{1}}{E}\right)=\frac{\frac{1}{2} \times \frac{20}{30}}{\frac{1}{2} \times \frac{20}{30}+\frac{1}{2} \times \frac{15}{20}}=\frac{\frac{2}{3}}{\frac{2}{3}+\frac{3}{4}}=\frac{8}{17}$

 

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