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:

• A. $\frac{4}{17}$
• B. $\frac{8}{17}$
• C. $\frac{2}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B

$\frac{8}{17}$

Explanation for correct options:

Step 1: Considering the elements of Box 1 and Box 2

Box 1 = { 1, 2, 3, 4, 5….30 }

Prime numbers in Box 1 = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }

Count of cards with prime numbers = 10

Total number of cards = 30

Count of cards with non-prime numbers = $30-10$

= $20$

Probability of Box 1 getting selected = $\frac{1}{2}$

Probability of non-prime numbers getting selected = $\frac{20}{30}$

= $\frac{2}{3}$

Box 2 = { 31, 32, 33, 34,….,50 }

Prime numbers in Box 2 = { 31, 37, 41, 43, 47 }

Count of prime numbers = 5

Total number of cards = 20

Number of non-prime cards = $20-5$

= $15$

Probability of Box 2 getting selected = $\frac{1}{2}$

Probability of non-prime numbers getting selected = $\frac{15}{20}$

Step 2: Applying Theorem

The probability of picking a non-prime card from Box 1 = $\frac{P(Box1)\times P(Non-Prime)}{\left\{P\right(Box1)\times P(Non-Prime1\left)\right\}+\left\{P\right(Box2)\times P(Non-Prime2\left)\right\}}$

= $\frac{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}}{({\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}})+({\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{4}})}$

= $\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{2}{3}}+{\displaystyle \frac{3}{4}}}$

= $\frac{2\times 12}{3(8+9)}$

= $\frac{8}{17}$

Therefore, the probability of selecting a non-prime card from Box 1 is $\frac{8}{17}$

Hence, Option (B) is the correct .