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 step by step, we need to analyze the given quadratic inequality and the probability of hitting the target in at most 'n' attempts.

### Step 1: Understand the Problem
The man has a 20% chance of hitting the target in one shot, which means:
- Probability of hitting the target (P(A)) = 0.2 = \( \frac{1}{5} \)
- Probability of missing the target (P(A')) = 1 - P(A) = 0.8 = \( \frac{4}{5} \)

### Step 2: Set Up the Probability Expression
We need to find the probability \( P \) of hitting the target in at most 'n' attempts. This can be expressed as:
\[ P = P(\text{hit in 1st attempt}) + P(\text{hit in 2nd attempt}) + \ldots + P(\text{hit in n-th attempt}) \]

The probability of hitting the target in at most 'n' attempts can be calculated using the formula for a geometric series:
\[ P = P(A) + P(A')P(A) + P(A')^2P(A) + \ldots + P(A')^{n-1}P(A) \]

This simplifies to:
\[ P = P(A) \left( 1 + P(A') + P(A')^2 + \ldots + P(A')^{n-1} \right) \]

### Step 3: Use the Formula for the Sum of a Geometric Series
The sum of the first 'n' terms of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

Here, \( a = 1 \) and \( r = P(A') = \frac{4}{5} \). Thus:
\[ P = \frac{1}{5} \cdot \frac{1 - \left(\frac{4}{5}\right)^n}{1 - \frac{4}{5}} \]
\[ P = \frac{1}{5} \cdot \frac{1 - \left(\frac{4}{5}\right)^n}{\frac{1}{5}} \]
\[ P = 1 - \left(\frac{4}{5}\right)^n \]

### Step 4: Set Up the Inequality
We are given the inequality:
\[ 20P^2 - 13P + 2 \leq 0 \]
Substituting \( P = 1 - \left(\frac{4}{5}\right)^n \):
\[ 20(1 - \left(\frac{4}{5}\right)^n)^2 - 13(1 - \left(\frac{4}{5}\right)^n) + 2 \leq 0 \]

### Step 5: Solve the Quadratic Inequality
Let \( x = \left(\frac{4}{5}\right)^n \). Then we rewrite the inequality:
\[ 20(1 - 2x + x^2) - 13(1 - x) + 2 \leq 0 \]
This simplifies to:
\[ 20 - 40x + 20x^2 - 13 + 13x + 2 \leq 0 \]
\[ 20x^2 - 27x + 9 \leq 0 \]

### Step 6: Find the Roots of the Quadratic
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 20, b = -27, c = 9 \):
\[ x = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 20 \cdot 9}}{2 \cdot 20} \]
\[ x = \frac{27 \pm \sqrt{729 - 720}}{40} \]
\[ x = \frac{27 \pm 3}{40} \]
This gives us:
\[ x_1 = \frac{30}{40} = \frac{3}{4}, \quad x_2 = \frac{24}{40} = \frac{3}{5} \]

### Step 7: Determine the Range of x
The inequality \( 20x^2 - 27x + 9 \leq 0 \) holds between the roots:
\[ \frac{3}{5} \leq x \leq \frac{3}{4} \]

### Step 8: Relate x back to n
Since \( x = \left(\frac{4}{5}\right)^n \):
1. For \( x = \frac{3}{5} \):
\[ \left(\frac{4}{5}\right)^n = \frac{3}{5} \]
Taking logarithms:
\[ n \log\left(\frac{4}{5}\right) = \log\left(\frac{3}{5}\right) \]
\[ n = \frac{\log\left(\frac{3}{5}\right)}{\log\left(\frac{4}{5}\right)} \]

2. For \( x = \frac{3}{4} \):
\[ \left(\frac{4}{5}\right)^n = \frac{3}{4} \]
Taking logarithms:
\[ n = \frac{\log\left(\frac{3}{4}\right)}{\log\left(\frac{4}{5}\right)} \]

### Step 9: Calculate Maximum n
Using numerical values:
- \( \log\left(\frac{3}{5}\right) \) and \( \log\left(\frac{3}{4}\right) \) can be approximated using a calculator.
- After calculation, we find that the maximum integer value of \( n \) satisfying the inequality is \( n = 2 \).

### Conclusion
The maximum value of \( n \) is:
\[ \boxed{2} \]