Mr. B has two fair 6-sided dice, one whose faces are numbered 1 to 6 and the second whose faces are numbered 3 to 8. Twice, he randomly picks one of dice (each dice equally likely) and rolls it. Given the sum of the resulting two rolls is 9, The probability he rolled same dice twice is `(m)/(n)` where m and n are relatively prime positive integers. Then the value of (m+n) is

Answer

To solve the problem step by step, we will calculate the required probabilities and use the concept of conditional probability.

### Step 1: Identify the Dice and Their Outcomes
Mr. B has two dice:
- **Die 1**: Faces numbered 1 to 6.
- **Die 2**: Faces numbered 3 to 8.

### Step 2: Calculate Possible Outcomes
Each die has 6 faces, so the total outcomes when rolling one die twice is:
- For Die 1: \(6 \times 6 = 36\) outcomes.
- For Die 2: \(6 \times 6 = 36\) outcomes.

### Step 3: Define Events
Let:
- \(A\): Event that the same die is rolled twice.
- \(B\): Event that the sum of the rolls is 9.

We need to find \(P(A|B)\), the probability that the same die was rolled given that the sum is 9.

### Step 4: Calculate \(P(A \cap B)\)
We need to find the cases where the sum equals 9 for the same die.

#### Case 1: Using Die 1
The pairs that give a sum of 9 are:
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)

Total favorable outcomes for Die 1: **4 outcomes**.

#### Case 2: Using Die 2
The pairs that give a sum of 9 are:
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)
- (7, 2)
- (8, 1)

Total favorable outcomes for Die 2: **4 outcomes**.

Thus, \(P(A \cap B) = \frac{4}{36} + \frac{4}{36} = \frac{8}{36} = \frac{2}{9}\).

### Step 5: Calculate \(P(B)\)
Now we need to find the total outcomes that give a sum of 9 regardless of the die used.

#### Case 1: Die 1 and Die 1
We already found **4 outcomes**.

#### Case 2: Die 2 and Die 2
We also found **4 outcomes**.

#### Case 3: Die 1 and Die 2
The pairs that give a sum of 9 are:
- (1, 8)
- (2, 7)
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)

Total favorable outcomes for this case: **6 outcomes**.

#### Case 4: Die 2 and Die 1
The pairs that give a sum of 9 are the same as above, so again **6 outcomes**.

Thus, total outcomes for \(B\):
\[ P(B) = \frac{4}{36} + \frac{4}{36} + \frac{6}{36} + \frac{6}{36} = \frac{20}{36} = \frac{5}{9}. \]

### Step 6: Calculate \(P(A|B)\)
Using the formula for conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}. \]

### Step 7: Find \(m+n\)
Here, \(m = 2\) and \(n = 5\). Therefore, \(m+n = 2 + 5 = 7\).

### Final Answer
The value of \(m+n\) is **7**.

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