Probability of throwing a 5 only on the first of two successive throws of an ordinary die is

Options

A. `1/36`
B. `5/36`
C. `25/36`
D. `1/6`

Answer

To solve the problem of finding the probability of throwing a 5 only on the first of two successive throws of an ordinary die, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Event**: We want to find the probability of getting a 5 on the first throw and not getting a 5 on the second throw.

2. **Favorable Outcomes for the First Throw**:
- The only favorable outcome for the first throw is getting a 5.
- Therefore, the number of favorable outcomes for the first throw = 1 (which is rolling a 5).

3. **Favorable Outcomes for the Second Throw**:
- For the second throw, we want to avoid rolling a 5. The possible outcomes are 1, 2, 3, 4, or 6.
- Thus, the number of favorable outcomes for the second throw = 5 (which are rolling 1, 2, 3, 4, or 6).

4. **Total Favorable Outcomes**:
- Since the first throw must be a 5 and the second throw must be one of the other five numbers, the total number of favorable outcomes = 1 (for the first throw) × 5 (for the second throw) = 5.

5. **Total Possible Outcomes**:
- Each die has 6 faces, and since we are throwing the die twice, the total number of possible outcomes = 6 (for the first throw) × 6 (for the second throw) = 36.

6. **Calculate the Probability**:
- The probability of the event is given by the formula:
\[ P(\text{throwing a 5 only on the first throw}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{5}{36}. \]

### Final Answer:
The probability of throwing a 5 only on the first of two successive throws of an ordinary die is \( \frac{5}{36} \).

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