The probability of a man hitting a target is `(2)/(5)` . He fires at the target k times (k, a given number) . Then the minimum k, so that the probability of hitting the targer at least once is more than `(7)/(10)`, is

Options

A. 3
B. 5
C. 2
D. 4

Answer

To solve the problem, we need to find the minimum value of \( k \) such that the probability of hitting the target at least once is greater than \( \frac{7}{10} \).

### Step-by-Step Solution:

1. **Identify the Probability of Hitting and Not Hitting the Target:**
- The probability of hitting the target, \( P(H) = \frac{2}{5} \).
- The probability of not hitting the target, \( P(NH) = 1 - P(H) = 1 - \frac{2}{5} = \frac{3}{5} \).

2. **Probability of Not Hitting the Target in \( k \) Shots:**
- The probability of not hitting the target in \( k \) shots is given by:
\[ P(NH \text{ in } k \text{ shots}) = \left( \frac{3}{5} \right)^k \]

3. **Probability of Hitting the Target at Least Once:**
- The probability of hitting the target at least once in \( k \) shots is:
\[ P(H \text{ at least once}) = 1 - P(NH \text{ in } k \text{ shots}) = 1 - \left( \frac{3}{5} \right)^k \]

4. **Set Up the Inequality:**
- We want this probability to be greater than \( \frac{7}{10} \):
\[ 1 - \left( \frac{3}{5} \right)^k > \frac{7}{10} \]

5. **Rearranging the Inequality:**
- Rearranging gives:
\[ \left( \frac{3}{5} \right)^k < 1 - \frac{7}{10} \]
- Simplifying the right side:
\[ 1 - \frac{7}{10} = \frac{3}{10} \]
- Thus, we have:
\[ \left( \frac{3}{5} \right)^k < \frac{3}{10} \]

6. **Taking Logarithms:**
- Taking logarithm on both sides:
\[ k \cdot \log\left( \frac{3}{5} \right) < \log\left( \frac{3}{10} \right) \]
- Since \( \log\left( \frac{3}{5} \right) \) is negative, we can reverse the inequality:
\[ k > \frac{\log\left( \frac{3}{10} \right)}{\log\left( \frac{3}{5} \right)} \]

7. **Calculating the Values:**
- Using logarithm values:
- \( \log(3) \approx 0.4771 \)
- \( \log(5) \approx 0.6990 \)
- \( \log(10) = 1 \)
- Therefore:
\[ \log\left( \frac{3}{10} \right) = \log(3) - \log(10) \approx 0.4771 - 1 = -0.5229 \]
\[ \log\left( \frac{3}{5} \right) = \log(3) - \log(5) \approx 0.4771 - 0.6990 = -0.2219 \]
- Now substituting:
\[ k > \frac{-0.5229}{-0.2219} \approx 2.36 \]

8. **Finding Minimum Integer Value of \( k \):**
- Since \( k \) must be an integer, we round up:
\[ k \geq 3 \]

### Conclusion:
The minimum value of \( k \) such that the probability of hitting the target at least once is greater than \( \frac{7}{10} \) is \( k = 3 \).