In a series of 7, T-20 matches between India and England, winning probability is `(2)/(5)` and `(3)/(5)` respectively for India and England. If probability of India winning n matches is highest then n is

Explanation

To find the value of \( n \) such that the probability of India winning \( n \) matches is highest in a series of 7 T20 matches against England, we can use the binomial probability formula.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The total number of matches \( n = 7 \).
- The probability of India winning a match \( p = \frac{2}{5} \).
- The probability of England winning a match \( q = \frac{3}{5} \).

2. **Binomial Probability Formula**:
The probability of India winning exactly \( k \) matches out of \( n \) matches is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

3. **Finding the Mode of the Distribution**:
The mode of a binomial distribution occurs at \( k \) values close to \( (n+1)p \). For our case:
\[ k = (n+1)p = (7+1) \cdot \frac{2}{5} = 8 \cdot \frac{2}{5} = \frac{16}{5} = 3.2 \]
Since \( k \) must be an integer, we consider \( k = 3 \) and \( k = 4 \).

4. **Calculating Probabilities for \( k = 3 \) and \( k = 4 \)**:
- For \( k = 3 \):
\[ P(X = 3) = \binom{7}{3} \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^{4} \]
- For \( k = 4 \):
\[ P(X = 4) = \binom{7}{4} \left(\frac{2}{5}\right)^4 \left(\frac{3}{5}\right)^{3} \]

5. **Calculating Binomial Coefficients**:
- \( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)
- \( \binom{7}{4} = \binom{7}{3} = 35 \)

6. **Calculating the Probabilities**:
- For \( k = 3 \):
\[ P(X = 3) = 35 \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^{4} = 35 \cdot \frac{8}{125} \cdot \frac{81}{625} = 35 \cdot \frac{648}{78125} \]
- For \( k = 4 \):
\[ P(X = 4) = 35 \left(\frac{2}{5}\right)^4 \left(\frac{3}{5}\right)^{3} = 35 \cdot \frac{16}{625} \cdot \frac{27}{125} = 35 \cdot \frac{432}{78125} \]

7. **Comparing Probabilities**:
- Calculate \( P(X = 3) \) and \( P(X = 4) \):
\[ P(X = 3) = \frac{35 \cdot 648}{78125} \quad \text{and} \quad P(X = 4) = \frac{35 \cdot 432}{78125} \]
- Since \( 648 > 432 \), we find that \( P(X = 3) > P(X = 4) \).

8. **Conclusion**:
The value of \( n \) for which the probability of India winning \( n \) matches is highest is:
\[ \boxed{3} \]