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Step-by-Step Solution Using Bayes' Theorem
Step 1: Identify the Events
Let:
• $B_1$: Event that Box I (with cards numbered 1 to 30) is selected.
• $B_2$: Event that Box II (with cards numbered 31 to 50) is selected.
• $E$: Event that the card drawn is a non-prime number.
Step 2: Determine the Probabilities of Selecting Each Box
Since one box is selected at random,
$P(B_1) = \tfrac{1}{2}$ and $P(B_2) = \tfrac{1}{2}.$
Step 3: Compute the Probability of Drawing a Non-Prime from Each Box
Box I (cards 1 to 30)
Prime numbers in Box I: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} (10 primes)
Therefore, total non-prime cards in Box I = 30 − 10 = 20.
$P(E \mid B_1) = \tfrac{\text{(number of non-primes in Box I)}}{\text{(total cards in Box I)}} = \tfrac{20}{30} = \tfrac{2}{3}.$
Box II (cards 31 to 50)
Prime numbers in Box II: {31, 37, 41, 43, 47} (5 primes)
Therefore, total non-prime cards in Box II = 20 − 5 = 15.
$P(E \mid B_2) = \tfrac{\text{(number of non-primes in Box II)}}{\text{(total cards in Box II)}} = \tfrac{15}{20} = \tfrac{3}{4}.$
Step 4: Use Bayes' Theorem to Find $P(B_1 \mid E)$
Bayes’ Theorem states:
$$
P(B_1 \mid E) = \frac{P(B_1)\, P(E \mid B_1)}{P(B_1)\, P(E \mid B_1) + P(B_2)\, P(E \mid B_2)}.
$$
Substituting the values:
$$
P(B_1 \mid E)
= \frac{\tfrac{1}{2} \times \tfrac{2}{3}}{\tfrac{1}{2} \times \tfrac{2}{3} + \tfrac{1}{2} \times \tfrac{3}{4}}
= \frac{\tfrac{2}{3}}{\tfrac{2}{3} + \tfrac{3}{4}}.
$$
Convert to a common denominator:
$$
\tfrac{2}{3} = \tfrac{8}{12},
\quad
\tfrac{3}{4} = \tfrac{9}{12}.
$$
So the denominator becomes $\tfrac{8}{12} + \tfrac{9}{12} = \tfrac{17}{12}.$
Therefore:
$$
P(B_1 \mid E)
= \frac{\tfrac{8}{12}}{\tfrac{17}{12}}
= \tfrac{8}{17}.
$$
Final Answer
The probability that the card was drawn from Box I given that it is a non-prime number is
$$
\boxed{\tfrac{8}{17}}.
$$
