If the expected number of sixes hit by a batsman on 
n
n balls is 4.24 and the variance for the same is 1.9928, then what is the probability of him hitting at least one six on any randomly selected 
n
n balls? Assume that the probability of him hitting a six on any ball is constant. (Enter the answer correct to 4 decimal places)

To solve this problem, we'll approach it using the relationship between the expected value, variance, and the binomial distribution.

### Step 1: Define the parameters
- Let \( p \) be the probability of the batsman hitting a six on any given ball.
- The expected number of sixes over \( n \) balls is given by the expected value \( E(X) \), where \( X \) is the number of sixes hit.
- The variance of the number of sixes hit is given by \( \text{Var}(X) \).

### Step 2: Use the given information
The problem provides:
- Expected value \( E(X) = np = 4.24 \)
- Variance \( \text{Var}(X) = np(1-p) = 1.9928 \)

### Step 3: Solve for \( p \) and \( n \)
We have two equations:
1. \( np = 4.24 \)
2. \( np(1-p) = 1.9928 \)

From the first equation:
\[
p = \frac{4.24}{n}
\]

Substitute this into the second equation:
\[
\frac{4.24}{n} \cdot n \left( 1 - \frac{4.24}{n} \right) = 1.9928
\]

This simplifies to:
\[
4.24 \left( 1 - \frac{4.24}{n} \right) = 1.9928
\]

Expanding and simplifying:
\[
4.24 - \frac{4.24^2}{n} = 1.9928
\]

\[
\frac{4.24^2}{n} = 4.24 - 1.9928 = 2.2472
\]

So:
\[
n = \frac{4.24^2}{2.2472}
\]

\[
n \approx 8
\]

Now, using \( n \approx 8 \), calculate \( p \):
\[
p = \frac{4.24}{8} = 0.53
\]

### Step 4: Calculate the probability of hitting at least one six
The probability of not hitting a six on any given ball is \( 1 - p \). Thus, the probability of not hitting a six on all \( n \) balls is \( (1 - p)^n \).

The probability of hitting at least one six is then:
\[
P(\text{at least one six}) = 1 - (1 - p)^n
\]

Substituting the values:
\[
P(\text{at least one six}) = 1 - (1 - 0.53)^8
\]

### Step 5: Compute the final probability
Let's calculate this value.

\[
P(\text{at least one six}) = 1 - (0.47)^8
\]
\[
P(\text{at least one six}) \approx 1 - 0.0006634
\]
\[
P(\text{at least one six}) \approx 0.9993366
\]

Thus, the probability of hitting at least one six on any randomly selected \( n \) balls is approximately **0.9993** (correct to 4 decimal places).

Would you like more details or have any questions? Here are some related questions:

1. How would the probability change if the expected number of sixes increased?
2. What would happen to the probability if the variance was larger?
3. How does the binomial distribution apply to this problem?
4. What is the significance of variance in determining the probability?
5. How would you interpret the probability of hitting zero sixes?

**Tip:** When solving problems involving binomial distributions, always start by relating the expected value and variance to find key parameters like \( n \) and \( p \).

Learn how to calculate the probability of a batsman hitting at least one six in cricket, given the expected number of sixes and variance. Step-by-step solution using probability theory and binomial distribution. Suitable for advanced high school students and beyond.

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