The chance of throwing an ace first only of two successive throws with 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 an ace (1) first only in two successive throws of an ordinary die, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
We need to find the probability of getting a 1 (ace) on the first throw and not getting a 1 on the second throw.

2. **Total Outcomes**:
An ordinary die has 6 faces, so the total number of possible outcomes when throwing a die is 6.

3. **Favorable Outcomes for the First Throw**:
For the first throw, we want to get a 1. There is only 1 favorable outcome (getting a 1).

\[ P(\text{1 on first throw}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{6} \]

4. **Favorable Outcomes for the Second Throw**:
For the second throw, we want to not get a 1. The favorable outcomes are 2, 3, 4, 5, and 6. Thus, there are 5 favorable outcomes.

\[ P(\text{not 1 on second throw}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{6} \]

5. **Combining the Probabilities**:
Since the events are independent (the outcome of the first throw does not affect the outcome of the second throw), we multiply the probabilities of the two events.

\[ P(\text{1 on first throw and not 1 on second throw}) = P(\text{1 on first throw}) \times P(\text{not 1 on second throw}) = \frac{1}{6} \times \frac{5}{6} \]

6. **Calculating the Final Probability**:
Now, we calculate the product:

\[ P = \frac{1 \times 5}{6 \times 6} = \frac{5}{36} \]

Thus, the probability of throwing an ace first only in two successive throws with an ordinary die is \(\frac{5}{36}\).

### Final Answer
\[ \text{The required probability is } \frac{5}{36}. \]