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. If the number is non-prime, the probability that it came from Box I is:
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. If the number is non-prime, the probability that it came from Box I is:
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Remember that 1 is not a prime number. Always include 1 when counting non-primes in the range [1, N].
Updated On: Jun 9, 2026
- \(\frac{4}{17} \)
- \(\frac{8}{17} \)
- \(\frac{2}{5} \)
- \(\frac{2}{3} \)
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The Correct Option is B
Solution and Explanation
Concept:
This problem requires Bayes' Theorem, which relates conditional probabilities: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$. Here, $A$ is the box selected and $B$ is the event of drawing a non-prime card.
Step 1: Determine the number of primes and non-primes in each box.
Box I (1-30): Primes are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$. There are 10 prime numbers.
Number of non-primes in Box I $= 30 - 10 = 20$.
Box II (31-50): Primes are $\{31, 37, 41, 43, 47\}$. There are 5 prime numbers.
Number of non-primes in Box II $= 20 - 5 = 15$.
Step 2: Calculate the probability of drawing a non-prime from each box.
Let $H_1$ be selecting Box I and $H_2$ be selecting Box II. $P(H_1) = 0.5$, $P(H_2) = 0.5$.
Let $E$ be the event of drawing a non-prime card.
$P(E|H_1) = \frac{20}{30} = \frac{2}{3}$.
$P(E|H_2) = \frac{15}{20} = \frac{3}{4}$.
Step 3: Apply Bayes' Theorem to find \(P(H_1|E)\).
\[ P(H_1|E) = \frac{P(H_1) \cdot P(E|H_1)}{P(H_1) \cdot P(E|H_1) + P(H_2) \cdot P(E|H_2)} \] \[ P(H_1|E) = \frac{0.5 \cdot (2/3)}{0.5 \cdot (2/3) + 0.5 \cdot (3/4)} \] The $0.5$ cancels out:
\[ P(H_1|E) = \frac{2/3}{2/3 + 3/4} = \frac{2/3}{(8+9)/12} = \frac{2/3}{17/12} = \frac{2}{3} \cdot \frac{12}{17} = \frac{8}{17} \] center minipage0.3
Probability = 8/17 minipage center
Step 1: Determine the number of primes and non-primes in each box.
Box I (1-30): Primes are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$. There are 10 prime numbers.
Number of non-primes in Box I $= 30 - 10 = 20$.
Box II (31-50): Primes are $\{31, 37, 41, 43, 47\}$. There are 5 prime numbers.
Number of non-primes in Box II $= 20 - 5 = 15$.
Step 2: Calculate the probability of drawing a non-prime from each box.
Let $H_1$ be selecting Box I and $H_2$ be selecting Box II. $P(H_1) = 0.5$, $P(H_2) = 0.5$.
Let $E$ be the event of drawing a non-prime card.
$P(E|H_1) = \frac{20}{30} = \frac{2}{3}$.
$P(E|H_2) = \frac{15}{20} = \frac{3}{4}$.
Step 3: Apply Bayes' Theorem to find \(P(H_1|E)\).
\[ P(H_1|E) = \frac{P(H_1) \cdot P(E|H_1)}{P(H_1) \cdot P(E|H_1) + P(H_2) \cdot P(E|H_2)} \] \[ P(H_1|E) = \frac{0.5 \cdot (2/3)}{0.5 \cdot (2/3) + 0.5 \cdot (3/4)} \] The $0.5$ cancels out:
\[ P(H_1|E) = \frac{2/3}{2/3 + 3/4} = \frac{2/3}{(8+9)/12} = \frac{2/3}{17/12} = \frac{2}{3} \cdot \frac{12}{17} = \frac{8}{17} \] center minipage0.3
Probability = 8/17 minipage center
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