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 :

Given:
Box I → cards numbered 1 to 30 (30 cards)
Box II → cards numbered 31 to 50 (20 cards)
A box is selected at random → probability of each box = $\dfrac{1}{2}$

Non-prime numbers in each box:
Box I (1–30): Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 → 10 primes.
Non-prime numbers = 30 − 10 = 20
(Including 1 as non-prime)

Box II (31–50): Prime numbers are 31, 37, 41, 43, 47 → 5 primes.
Non-prime numbers = 20 − 5 = 15

Let
A = “card drawn from Box I”
B = “card drawn from Box II”
N = “number on the card is non-prime”

We need $P(A|N)$.

Using Bayes’ theorem:
P(A|N) = \frac{P(A)P(N|A)}{P(A)P(N|A) + P(B)P(N|B)} \]
Now substitute:
\[ P(A) = P(B) = \frac{1}{2}, \quad \] \[ P(N|A) \frac{20}{30} = \frac{2}{3}, \quad \] \[ P(N|B) = \frac{15}{20} = \frac{3}{4} \]
\[ P(A|N) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{3}{4}} \] \[= \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} \] \[= \frac{\frac{1}{3}}{\frac{17}{24}}\] \[= \frac{8}{17} \]

Final Answer: $\boxed{\dfrac{8}{17}}$