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 number of matches India is most likely to win in a series of 7 T20 matches against England, we can use the binomial probability formula. Here’s a step-by-step solution:

### Step 1: Identify the probabilities
The probability of India winning a match (p) is \( \frac{2}{5} \) and the probability of England winning a match (q) is \( \frac{3}{5} \).

### Step 2: Define the binomial distribution
In a binomial distribution, the probability of winning \( n \) matches out of \( N \) total matches is given by the formula:
\[ P(X = n) = \binom{N}{n} p^n q^{N-n} \]
where:
- \( N = 7 \) (total matches)
- \( p = \frac{2}{5} \) (probability of India winning)
- \( q = \frac{3}{5} \) (probability of England winning)

### Step 3: Find the probability for each possible number of wins
We need to calculate the probability \( P(X = n) \) for \( n = 0, 1, 2, \ldots, 7 \).

### Step 4: Calculate the probabilities
Using the formula, we calculate \( P(X = n) \) for each \( n \):

1. **For \( n = 0 \)**:
\[ P(X = 0) = \binom{7}{0} \left(\frac{2}{5}\right)^0 \left(\frac{3}{5}\right)^7 = 1 \cdot 1 \cdot \left(\frac{2187}{78125}\right) = \frac{2187}{78125} \]

2. **For \( n = 1 \)**:
\[ P(X = 1) = \binom{7}{1} \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^6 = 7 \cdot \left(\frac{2}{5}\right) \cdot \left(\frac{729}{15625}\right) = \frac{10206}{78125} \]

3. **For \( n = 2 \)**:
\[ P(X = 2) = \binom{7}{2} \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^5 = 21 \cdot \left(\frac{4}{25}\right) \cdot \left(\frac{243}{3125}\right) = \frac{20412}{78125} \]

4. **For \( n = 3 \)**:
\[ P(X = 3) = \binom{7}{3} \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^4 = 35 \cdot \left(\frac{8}{125}\right) \cdot \left(\frac{81}{625}\right) = \frac{22680}{78125} \]

5. **For \( n = 4 \)**:
\[ P(X = 4) = \binom{7}{4} \left(\frac{2}{5}\right)^4 \left(\frac{3}{5}\right)^3 = 35 \cdot \left(\frac{16}{625}\right) \cdot \left(\frac{27}{125}\right) = \frac{18900}{78125} \]

6. **For \( n = 5 \)**:
\[ P(X = 5) = \binom{7}{5} \left(\frac{2}{5}\right)^5 \left(\frac{3}{5}\right)^2 = 21 \cdot \left(\frac{32}{3125}\right) \cdot \left(\frac{9}{25}\right) = \frac{6048}{78125} \]

7. **For \( n = 6 \)**:
\[ P(X = 6) = \binom{7}{6} \left(\frac{2}{5}\right)^6 \left(\frac{3}{5}\right)^1 = 7 \cdot \left(\frac{64}{15625}\right) \cdot \left(\frac{3}{5}\right) = \frac{2688}{78125} \]

8. **For \( n = 7 \)**:
\[ P(X = 7) = \binom{7}{7} \left(\frac{2}{5}\right)^7 \left(\frac{3}{5}\right)^0 = 1 \cdot \left(\frac{128}{78125}\right) \cdot 1 = \frac{128}{78125} \]

### Step 5: Compare the probabilities
Now we compare the probabilities calculated above to find the maximum value:
- \( P(X = 0) = \frac{2187}{78125} \)
- \( P(X = 1) = \frac{10206}{78125} \)
- \( P(X = 2) = \frac{20412}{78125} \)
- \( P(X = 3) = \frac{22680}{78125} \) (highest)
- \( P(X = 4) = \frac{18900}{78125} \)
- \( P(X = 5) = \frac{6048}{78125} \)
- \( P(X = 6) = \frac{2688}{78125} \)
- \( P(X = 7) = \frac{128}{78125} \)

### Conclusion
The highest probability occurs when \( n = 3 \). Therefore, the number of matches India is most likely to win is **3**.

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