1. Define events

Let:

• $B_1$ = card drawn from Box I
• $B_2$ = card drawn from Box II
• $N$ = drawn card has a non-prime number

Since a box is selected at random, $P(B_1)=P(B_2)=\frac12.$

We need: $P(B_1\mid N).$

Using Bayes' theorem, $P(B_1\mid N)=\frac{P(B_1)P(N\mid B_1)}{P(B_1)P(N\mid B_1)+P(B_2)P(N\mid B_2)}.$

2. Find non-prime numbers in Box I

Box I contains numbers $1$ to $30$.

Prime numbers from $1$ to $30$ are: $2,3,5,7,11,13,17,19,23,29$ So number of primes $=10$.

Hence number of non-primes in Box I: $30-10=20.$ Therefore, $P(N\mid B_1)=\frac{20}{30}=\frac23.$

3. Find non-prime numbers in Box II

Box II contains numbers $31$ to $50$, total $20$ numbers.

Primes from $31$ to $50$ are: $31,37,41,43,47$ So number of primes $=5$.

Hence number of non-primes in Box II: $20-5=15.$ Therefore, $P(N\mid B_2)=\frac{15}{20}=\frac34.$

4. Apply Bayes' theorem

$P(B_1\mid N)=\frac{\frac12\cdot\frac23}{\frac12\cdot\frac23+\frac12\cdot\frac34}.$

Cancel $\frac12$ from numerator and denominator: $P(B_1\mid N)=\frac{\frac23}{\frac23+\frac34}.$

Now, $\frac23+\frac34=\frac{8+9}{12}=\frac{17}{12}.$

So, P(B_1\mid N)=\frac{\frac23}{\frac{17}{12}}=rac23\cdot\frac{12}{17}=\frac{8}{17}.

5. Compare with options

$\frac{8}{17}$ which matches Option A.

6. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

So they agree.