Three dice, red, blue and green in colour are rolled together. Let B be the event that sum of the numbers shown up is 7. Let A be the event that the red die shows 1. The conditional probability of the event A given B, P(A|B) is

Options

A. `2/7`
B. `1/6`
C. `1/7`
D. `1/3`

Answer

To solve the problem step by step, we will calculate 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: Define the events
- Let \( R \) be the outcome of the red die.
- Let \( B \) be the outcome of the blue die.
- Let \( G \) be the outcome of the green die.

We need to find \( P(A|B) \).

### Step 2: Use the formula for conditional probability
The formula for conditional probability is given by:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

### Step 3: Calculate \( P(B) \)
To find \( P(B) \), we need to determine the total number of outcomes that result in the sum of the three dice being 7.

The possible combinations of \( (R, B, G) \) that sum to 7 are:
1. \( (1, 1, 5) \)
2. \( (1, 2, 4) \)
3. \( (1, 3, 3) \)
4. \( (2, 2, 3) \)
5. \( (2, 3, 2) \)
6. \( (3, 2, 2) \)
7. \( (1, 5, 1) \)
8. \( (5, 1, 1) \)
9. \( (4, 2, 1) \)
10. \( (4, 1, 2) \)
11. \( (3, 3, 1) \)
12. \( (3, 1, 3) \)
13. \( (1, 4, 2) \)
14. \( (2, 4, 1) \)

After considering all permutations, we find that there are 15 favorable outcomes that lead to a sum of 7.

The total number of outcomes when rolling three dice is \( 6 \times 6 \times 6 = 216 \).

Thus,

\[ P(B) = \frac{15}{216} = \frac{5}{72} \]

### Step 4: Calculate \( P(A \cap B) \)
Next, we need to find \( P(A \cap B) \), which is the probability that the red die shows 1 and the sum of the three dice is 7.

If the red die \( R = 1 \), then we need \( B + G = 6 \). The combinations for \( (B, G) \) that satisfy this are:
1. \( (1, 5) \)
2. \( (2, 4) \)
3. \( (3, 3) \)
4. \( (4, 2) \)
5. \( (5, 1) \)

This gives us 5 favorable outcomes.

Thus,

\[ P(A \cap B) = \frac{5}{216} \]

### Step 5: Calculate \( P(A|B) \)
Now we can substitute \( P(A \cap B) \) and \( P(B) \) into the conditional probability formula:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{216}}{\frac{15}{216}} = \frac{5}{15} = \frac{1}{3} \]

### Final Answer
The conditional probability \( P(A|B) \) is \( \frac{1}{3} \).

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