In a series of 5 T-20 matches between India and Australia, winning probability of India is `3/5` and of Australia is `2/5` . Then find the sum of mean and variance for India to win.

Options

A. `15/5`
B. `4/5`
C. `21/5`
D. `9/5`

Explanation

To solve the problem, we need to calculate the mean and variance of the number of matches India is expected to win in a series of 5 T-20 matches against Australia, given the probabilities of winning.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of matches (n) = 5
- Probability of India winning (p) = \( \frac{3}{5} \)
- Probability of Australia winning (q) = \( \frac{2}{5} \)

2. **Calculate the mean (μ):**
The mean (μ) for a binomial distribution is given by the formula:
\[ \mu = n \times p \]
Substituting the values:
\[ \mu = 5 \times \frac{3}{5} = 3 \]

3. **Calculate the variance (σ²):**
The variance (σ²) for a binomial distribution is given by the formula:
\[ \sigma^2 = n \times p \times q \]
Substituting the values:
\[ \sigma^2 = 5 \times \frac{3}{5} \times \frac{2}{5} \]
Simplifying:
\[ \sigma^2 = 5 \times \frac{3 \times 2}{5 \times 5} = \frac{6}{5} \]

4. **Calculate the sum of mean and variance:**
Now, we need to find the sum of the mean and variance:
\[ \text{Sum} = \mu + \sigma^2 \]
Substituting the values:
\[ \text{Sum} = 3 + \frac{6}{5} \]
To add these, we convert 3 into a fraction:
\[ 3 = \frac{15}{5} \]
Now adding:
\[ \text{Sum} = \frac{15}{5} + \frac{6}{5} = \frac{21}{5} \]

### Final Answer:
The sum of the mean and variance for India to win is \( \frac{21}{5} \).

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