A man firing a distant target has `20%` chance of hitting the target in one shot. If P be the probability of hitting the target in atmost 'n' attempts where `20P^(2)-13P+2 le0`. then maximum value of n is.

Explanation

To solve the problem, we need to find the maximum value of \( n \) such that the probability \( P \) of hitting a target in at most \( n \) attempts satisfies the inequality:

\[ 20P^2 - 13P + 2 \leq 0 \]

### Step 1: Solve the quadratic inequality

First, we need to solve the quadratic equation:

\[ 20P^2 - 13P + 2 = 0 \]

Using the quadratic formula \( P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 20 \), \( b = -13 \), and \( c = 2 \):

\[ P = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 20 \cdot 2}}{2 \cdot 20} \]

Calculating the discriminant:

\[ (-13)^2 - 4 \cdot 20 \cdot 2 = 169 - 160 = 9 \]

Now substituting back into the formula:

\[ P = \frac{13 \pm 3}{40} \]

This gives us two solutions:

\[ P_1 = \frac{16}{40} = \frac{2}{5}, \quad P_2 = \frac{10}{40} = \frac{1}{4} \]

### Step 2: Determine the range of \( P \)

The quadratic opens upwards (since \( a = 20 > 0 \)), so the inequality \( 20P^2 - 13P + 2 \leq 0 \) holds between the roots:

\[ \frac{1}{4} \leq P \leq \frac{2}{5} \]

### Step 3: Find the probability of hitting the target

Given that the man has a 20% chance of hitting the target in one shot, we have:

\[ p_H = \frac{1}{5} \]

The probability of missing the target in one shot is:

\[ p_H' = 1 - p_H = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 4: Probability of hitting the target in at most \( n \) attempts

The probability of hitting the target in at most \( n \) attempts can be expressed as:

\[ P = 1 - \left( \frac{4}{5} \right)^n \]

### Step 5: Set up the inequality

We need to satisfy:

\[ \frac{1}{4} \leq 1 - \left( \frac{4}{5} \right)^n \leq \frac{2}{5} \]

### Step 6: Solve the inequalities

1. **For the left inequality:**

\[ \frac{1}{4} \leq 1 - \left( \frac{4}{5} \right)^n \]

This simplifies to:

\[ \left( \frac{4}{5} \right)^n \leq \frac{3}{4} \]

Taking logarithm on both sides:

\[ n \log\left( \frac{4}{5} \right) \leq \log\left( \frac{3}{4} \right) \]

Since \( \log\left( \frac{4}{5} \right) < 0 \), we reverse the inequality:

\[ n \geq \frac{\log\left( \frac{3}{4} \right)}{\log\left( \frac{4}{5} \right)} \]

Calculating the values:

\[ \log\left( \frac{3}{4} \right) \approx -0.1249, \quad \log\left( \frac{4}{5} \right) \approx -0.0969 \]

Thus:

\[ n \geq \frac{-0.1249}{-0.0969} \approx 1.29 \]

So \( n \geq 2 \).

2. **For the right inequality:**

\[ 1 - \left( \frac{4}{5} \right)^n \leq \frac{2}{5} \]

This simplifies to:

\[ \left( \frac{4}{5} \right)^n \geq \frac{3}{5} \]

Taking logarithm on both sides:

\[ n \log\left( \frac{4}{5} \right) \geq \log\left( \frac{3}{5} \right) \]

Reversing the inequality:

\[ n \leq \frac{\log\left( \frac{3}{5} \right)}{\log\left( \frac{4}{5} \right)} \]

Calculating the values:

\[ \log\left( \frac{3}{5} \right) \approx -0.2218 \]

Thus:

\[ n \leq \frac{-0.2218}{-0.0969} \approx 2.29 \]

So \( n \leq 2 \).

### Conclusion

Combining both inequalities, we find:

\[ 2 \leq n \leq 2 \]

Thus, the maximum value of \( n \) is:

\[ \boxed{2} \]