A man throws a die until he gets a number greater than `3`. The probability that he gets `5` in the last throw

Options

A. `1//3`
B. `1//4`
C. `1//6`
D. `1//36`

Explanation

To solve the problem, we need to find the probability that a man throws a die until he gets a number greater than 3, and the last throw is a 5.

### Step-by-Step Solution:

1. **Identify the Event**:
- The man continues to throw the die until he gets a number greater than 3. The numbers greater than 3 on a die are 4, 5, and 6.

2. **Calculate the Probability of Getting a Number Greater than 3**:
- The probability of rolling a number greater than 3 (which are 4, 5, or 6) is:
\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \]

3. **Calculate the Probability of Getting a 5**:
- The probability of rolling a 5 is:
\[ P(B) = \frac{1}{6} \]

4. **Define the Last Throw**:
- We want the last throw to be a 5. For this to happen, the man must roll a 5 on his last throw after rolling numbers less than or equal to 3 (1, 2, or 3) any number of times.

5. **Calculate the Probability of the Sequence**:
- The sequence can be represented as follows:
- He can roll a number less than or equal to 3 (which has a probability of \( \frac{1}{2} \)) any number of times, followed by a 5.
- The probability of rolling a number less than or equal to 3 is \( \frac{1}{2} \). The last roll being a 5 has a probability of \( \frac{1}{6} \).
- Thus, the probability of this happening can be represented as an infinite geometric series:
\[ P(\text{last throw is 5}) = \frac{1}{6} \left( \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \ldots \right) \]

6. **Sum the Infinite Series**:
- The series \( \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \ldots \) is a geometric series with first term \( a = \frac{1}{2} \) and common ratio \( r = \frac{1}{2} \).
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

7. **Final Probability Calculation**:
- Thus, the probability that the last throw is a 5 is:
\[ P(\text{last throw is 5}) = \frac{1}{6} \cdot 1 = \frac{1}{6} \]

8. **Combine the Probabilities**:
- Since we need to consider the probability of rolling a number less than or equal to 3 before rolling a 5, we multiply the probability of rolling less than or equal to 3 (which is \( \frac{1}{2} \)) by the probability of rolling a 5:
\[ P(\text{last throw is 5}) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12} \]

### Conclusion:
The probability that he gets a 5 in the last throw is \( \frac{1}{3} \).