A die is tossed 5 times. Getting and odd number is cosidered a success. Then, the variance of distribution of success, is

Options

A. `(8)/(3)`
B. `(3)/(8)`
C. `(4)/(5)`
D. `(5)/(4)`

Explanation

To solve the problem of finding the variance of the distribution of successes when a die is tossed 5 times, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Experiment**: We are tossing a die 5 times. Each toss can result in either a success (getting an odd number) or a failure (getting an even number).

2. **Define Success and Failure**:
- Success: Getting an odd number (1, 3, or 5).
- Failure: Getting an even number (2, 4, or 6).

3. **Determine the Total Outcomes**: A standard die has 6 faces, so the total number of outcomes when rolling a die is 6.

4. **Calculate the Probability of Success (p)**:
- There are 3 odd numbers (1, 3, 5) out of 6 possible outcomes.
- Therefore, the probability of success (p) is:
\[ p = \frac{\text{Number of odd outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \]

5. **Calculate the Probability of Failure (q)**:
- The probability of failure (q) is given by:
\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Determine the Number of Trials (n)**:
- Since the die is tossed 5 times, we have \( n = 5 \).

7. **Use the Variance Formula for Binomial Distribution**:
- The variance \( \sigma^2 \) of a binomial distribution is given by the formula:
\[ \sigma^2 = n \cdot p \cdot q \]

8. **Substitute the Values into the Variance Formula**:
- Now, substituting the values we have:
\[ \sigma^2 = 5 \cdot \frac{1}{2} \cdot \frac{1}{2} \]
- Simplifying this gives:
\[ \sigma^2 = 5 \cdot \frac{1}{4} = \frac{5}{4} \]

9. **Final Answer**:
- The variance of the distribution of successes when a die is tossed 5 times is:
\[ \frac{5}{4} \]