Two unbiased die are rolled together . Let A={(a,b): a+b=11} and B= { (a,b): a `ne` 5}, then P(A|B) equals

Options

A. `1//30`
B. `1//15`
C. `2//15`
D. `5//6`

Answer

To solve the problem, we need to find \( P(A|B) \), which is the probability of event A given event B.

### Step 1: Define the events A and B

- **Event A**: The set of outcomes where the sum of the numbers on the two dice equals 11.
- Possible outcomes for A:
- (5, 6)
- (6, 5)
- Therefore, \( A = \{ (5, 6), (6, 5) \} \).

- **Event B**: The set of outcomes where the first die is not equal to 5.
- Possible outcomes for B:
- All combinations except those where the first die is 5:
- (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
- (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
- (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
- (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
- (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
- Total outcomes in B: 30 outcomes (since there are 36 total outcomes from rolling two dice, and 6 outcomes where the first die is 5).

### Step 2: Calculate \( P(A \cap B) \)

- **Intersection of A and B**: We need to find outcomes that are in both A and B.
- From A, we have (5, 6) and (6, 5).
- Only (6, 5) is in B because (5, 6) has a first die of 5 which is excluded in B.
- Therefore, \( A \cap B = \{ (6, 5) \} \).
- Number of outcomes in \( A \cap B = 1 \).

### Step 3: Calculate \( P(B) \)

- The probability of event B is given by the number of favorable outcomes in B divided by the total outcomes.
- Total outcomes when rolling two dice = 36.
- Number of outcomes in B = 30.
- Therefore, \( P(B) = \frac{30}{36} = \frac{5}{6} \).

### Step 4: Calculate \( P(A|B) \)

- The formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
- We found \( P(A \cap B) = \frac{1}{36} \) and \( P(B) = \frac{30}{36} \).
- Now, substituting the values:
\[ P(A|B) = \frac{\frac{1}{36}}{\frac{30}{36}} = \frac{1}{30} \]

### Final Answer

Thus, \( P(A|B) = \frac{1}{30} \).

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