A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained. Then find the probability that 5 comes before 7.

Options

A. `2//5
B. `3//5`
C. `4//5`
D. None of these

Explanation

To solve the problem of finding the probability that a sum of 5 comes before a sum of 7 when rolling a pair of unbiased dice, we can follow these steps:

### Step 1: Identify the possible outcomes
When rolling two dice, the total number of outcomes is \(6 \times 6 = 36\).

### Step 2: Calculate the probability of rolling a sum of 5
The combinations that yield a sum of 5 are:
1. (1, 4)
2. (2, 3)
3. (3, 2)
4. (4, 1)

Thus, there are 4 outcomes that result in a sum of 5. Therefore, the probability \(P(A)\) of rolling a sum of 5 is:
\[ P(A) = \frac{4}{36} = \frac{1}{9} \]

### Step 3: Calculate the probability of rolling a sum of 7
The combinations that yield a sum of 7 are:
1. (1, 6)
2. (2, 5)
3. (3, 4)
4. (4, 3)
5. (5, 2)
6. (6, 1)

Thus, there are 6 outcomes that result in a sum of 7. Therefore, the probability \(P(B)\) of rolling a sum of 7 is:
\[ P(B) = \frac{6}{36} = \frac{1}{6} \]

### Step 4: Calculate the probability of neither sum occurring
The probability of neither a sum of 5 nor a sum of 7 occurring is calculated by subtracting the probabilities of A and B from 1.

The total probability of rolling either a sum of 5 or 7 is:
\[ P(A \cup B) = P(A) + P(B) = \frac{1}{9} + \frac{1}{6} \]

To add these fractions, we need a common denominator:
\[ P(A \cup B) = \frac{2}{18} + \frac{3}{18} = \frac{5}{18} \]

Thus, the probability of neither sum occurring is:
\[ P(C) = 1 - P(A \cup B) = 1 - \frac{5}{18} = \frac{13}{18} \]

### Step 5: Calculate the probability that 5 comes before 7
To find the probability that a sum of 5 occurs before a sum of 7, we can use the formula:
\[ P(A \text{ before } B) = \frac{P(A)}{P(A) + P(B)} \]

Substituting the values we calculated:
\[ P(A \text{ before } B) = \frac{P(A)}{P(A) + P(B)} = \frac{\frac{1}{9}}{\frac{1}{9} + \frac{1}{6}} \]

Finding a common denominator for the denominator:
\[ P(A \text{ before } B) = \frac{\frac{1}{9}}{\frac{2}{18} + \frac{3}{18}} = \frac{\frac{1}{9}}{\frac{5}{18}} = \frac{1}{9} \times \frac{18}{5} = \frac{2}{5} \]

### Final Answer
Thus, the probability that a sum of 5 comes before a sum of 7 is:
\[ \boxed{\frac{2}{5}} \]