To solve the problem, we need to find the conditional probability \( P(A|B) \), where:
- \( A \) is the event that the red die shows 1.
- \( B \) is the event that the sum of the numbers shown on the three dice is 7.
### Step 1: Calculate \( P(A) \)
The event \( A \) occurs when the red die shows 1. The other two dice (blue and green) can show any number from 1 to 6.
The total number of outcomes when rolling three dice is \( 6 \times 6 \times 6 = 216 \).
When the red die shows 1, the blue and green dice can take any of the 6 values each. Therefore, the number of favorable outcomes for event \( A \) is:
\[
\text{Favorable outcomes for } A = 6 \times 6 = 36
\]
Thus, the probability of \( A \) is:
\[
P(A) = \frac{\text{Favorable outcomes for } A}{\text{Total outcomes}} = \frac{36}{216} = \frac{1}{6}
\]
### Step 2: Calculate \( P(B) \)
Next, we need to find the probability of event \( B \), which is the event that the sum of the numbers on the three dice is 7.
We can list the combinations that give a sum of 7:
1. (1, 1, 5)
2. (1, 2, 4)
3. (1, 3, 3)
4. (2, 2, 3)
Now, we need to count the arrangements for each combination:
- For (1, 1, 5): There are \( \frac{3!}{2!1!} = 3 \) arrangements.
- For (1, 2, 4): There are \( 3! = 6 \) arrangements.
- For (1, 3, 3): There are \( \frac{3!}{2!1!} = 3 \) arrangements.
- For (2, 2, 3): There are \( \frac{3!}{2!1!} = 3 \) arrangements.
Now we sum these arrangements:
\[
\text{Total arrangements for } B = 3 + 6 + 3 + 3 = 15
\]
Thus, the probability of \( B \) is:
\[
P(B) = \frac{\text{Favorable outcomes for } B}{\text{Total outcomes}} = \frac{15}{216}
\]
### Step 3: Calculate \( P(A \cap B) \)
Now we need to find the intersection \( P(A \cap B) \), which is the probability that the red die shows 1 and the sum of the three dice is 7.
If the red die shows 1, we need the blue and green dice to sum to \( 7 - 1 = 6 \). The pairs that sum to 6 are:
1. (1, 5)
2. (2, 4)
3. (3, 3)
4. (4, 2)
5. (5, 1)
Counting the arrangements:
- (1, 5): 2 arrangements (1, 5) and (5, 1)
- (2, 4): 2 arrangements (2, 4) and (4, 2)
- (3, 3): 1 arrangement (3, 3)
Thus, the total arrangements for \( A \cap B \) is:
\[
\text{Total arrangements for } A \cap B = 2 + 2 + 1 = 5
\]
Therefore, the probability of \( A \cap B \) is:
\[
P(A \cap B) = \frac{5}{216}
\]
### Step 4: Calculate \( P(A|B) \)
Now we can find the conditional probability \( P(A|B) \):
\[
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{216}}{\frac{15}{216}} = \frac{5}{15} = \frac{1}{3}
\]
### Final Answer
Thus, the conditional probability \( P(A|B) \) is:
\[
\boxed{\frac{1}{3}}
\]
