On a toss of two dice, A throws a total of 5, then the probability that he will throw another 5 before the throws 7, is

Options

A. `(1)/(9)`
B. `(1)/(6)`
C. `(2)/(5)`
D. `(5)/(36)`.

Answer

To solve the problem, we need to find the probability that A will throw another 5 before he throws a 7 after already throwing a 5 with two dice.

### Step-by-Step Solution:

1. **Identify the Outcomes for 5 and 7**:
- The possible outcomes to get a sum of 5 when rolling two dice are:
- (1, 4)
- (2, 3)
- (3, 2)
- (4, 1)
- Therefore, there are **4 ways** to get a sum of 5.
- The possible outcomes to get a sum of 7 are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)
- Therefore, there are **6 ways** to get a sum of 7.

2. **Calculate Total Outcomes**:
- The total number of outcomes when rolling two dice is \(6 \times 6 = 36\).

3. **Calculate the Probability of Getting 5 or 7**:
- The probability of rolling a 5 in one throw is:
\[ P(5) = \frac{4}{36} = \frac{1}{9} \]
- The probability of rolling a 7 in one throw is:
\[ P(7) = \frac{6}{36} = \frac{1}{6} \]

4. **Calculate the Probability of Neither 5 nor 7**:
- The probability of rolling neither a 5 nor a 7 is:
\[ P(\text{neither 5 nor 7}) = 1 - P(5) - P(7) = 1 - \frac{1}{9} - \frac{1}{6} \]
- To calculate this, we need a common denominator. The least common multiple of 9 and 6 is 18.
- Convert \(P(5)\) and \(P(7)\):
\[ P(5) = \frac{2}{18}, \quad P(7) = \frac{3}{18} \]
- Therefore,
\[ P(\text{neither 5 nor 7}) = 1 - \left(\frac{2}{18} + \frac{3}{18}\right) = 1 - \frac{5}{18} = \frac{13}{18} \]

5. **Set Up the Probability Series**:
- We want to find the probability of rolling a 5 before rolling a 7. This can be modeled as a geometric series.
- The probability of rolling a 5 on the first attempt is \(\frac{1}{9}\).
- The probability of rolling neither a 5 nor a 7 is \(\frac{13}{18}\).
- The probability of rolling a 5 on the second attempt (after rolling neither 5 nor 7 first) is \(\frac{13}{18} \times \frac{1}{9}\).
- This continues indefinitely, forming a series.

6. **Sum the Infinite Series**:
- The series can be represented as:
\[ S = \frac{1}{9} + \left(\frac{13}{18} \times \frac{1}{9}\right) + \left(\frac{13}{18}\right)^2 \times \frac{1}{9} + \ldots \]
- This is a geometric series where the first term \(a = \frac{1}{9}\) and the common ratio \(r = \frac{13}{18}\).
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{1}{9}}{1 - \frac{13}{18}} = \frac{\frac{1}{9}}{\frac{5}{18}} = \frac{1}{9} \times \frac{18}{5} = \frac{2}{5} \]

### Final Answer:
The probability that A will throw another 5 before he throws a 7 is \(\frac{2}{5}\).