To solve the problem step by step, we will calculate the probability that a card drawn is from Box 1 given that the card is non-prime.
### Step 1: Define the events
Let:
- \( B_1 \): the event that Box 1 is selected.
- \( B_2 \): the event that Box 2 is selected.
- \( E \): the event that the drawn card is non-prime.
Since there are two boxes, the probability of selecting either box is:
\[
P(B_1) = P(B_2) = \frac{1}{2}
\]
### Step 2: Identify non-prime numbers in Box 1
Box 1 contains cards numbered from 1 to 30. The prime numbers in this range are:
- 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Thus, the non-prime numbers in Box 1 are:
- 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30
Counting these, we find there are 20 non-prime numbers in Box 1.
### Step 3: Identify non-prime numbers in Box 2
Box 2 contains cards numbered from 31 to 50. The prime numbers in this range are:
- 31, 37, 41, 43, 47
Thus, the non-prime numbers in Box 2 are:
- 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50
Counting these, we find there are 15 non-prime numbers in Box 2.
### Step 4: Calculate the total probability of drawing a non-prime card
Using the law of total probability, we can calculate \( P(E) \):
\[
P(E) = P(B_1) \cdot P(E|B_1) + P(B_2) \cdot P(E|B_2)
\]
Where:
- \( P(E|B_1) = \frac{20}{30} = \frac{2}{3} \) (non-prime cards from Box 1)
- \( P(E|B_2) = \frac{15}{20} = \frac{3}{4} \) (non-prime cards from Box 2)
Substituting the values, we get:
\[
P(E) = \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{2} \cdot \frac{3}{4}
\]
\[
P(E) = \frac{1}{3} + \frac{3}{8}
\]
To add these fractions, we find a common denominator (24):
\[
P(E) = \frac{8}{24} + \frac{9}{24} = \frac{17}{24}
\]
### Step 5: Calculate the required probability \( P(B_1|E) \)
Using Bayes' theorem:
\[
P(B_1|E) = \frac{P(E|B_1) \cdot P(B_1)}{P(E)}
\]
Substituting the known values:
\[
P(B_1|E) = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{17}{24}}
\]
\[
P(B_1|E) = \frac{\frac{1}{3}}{\frac{17}{24}} = \frac{1}{3} \cdot \frac{24}{17} = \frac{8}{17}
\]
### Final Answer
The probability that the card came from Box 1 given that it is non-prime is:
\[
\boxed{\frac{8}{17}}
\]
