If the probability of hitting a target by a shooter, in any shot is 1/3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than `(5)/(6)` is

Options

A. 6
B. 5
C. 4
D. 3

Answer

To solve the problem, we need to determine the minimum number of independent shots required by a shooter such that the probability of hitting a target at least once is greater than \( \frac{5}{6} \).

### Step-by-Step Solution:

1. **Identify the Probability of Hitting and Missing the Target**:
- The probability of hitting the target in any shot is given as \( P(H) = \frac{1}{3} \).
- Therefore, the probability of missing the target is \( P(M) = 1 - P(H) = 1 - \frac{1}{3} = \frac{2}{3} \).

2. **Define the Probability of Hitting at Least Once**:
- The probability of hitting the target at least once in \( n \) shots can be expressed as:
\[ P(\text{at least one hit}) = 1 - P(\text{no hits in } n \text{ shots}) \]
- The probability of missing the target in all \( n \) shots is:
\[ P(\text{no hits}) = \left( \frac{2}{3} \right)^n \]
- Therefore, we have:
\[ P(\text{at least one hit}) = 1 - \left( \frac{2}{3} \right)^n \]

3. **Set Up the Inequality**:
- We want the probability of hitting at least once to be greater than \( \frac{5}{6} \):
\[ 1 - \left( \frac{2}{3} \right)^n > \frac{5}{6} \]
- Rearranging this gives:
\[ \left( \frac{2}{3} \right)^n < \frac{1}{6} \]

4. **Take Logarithms**:
- To solve for \( n \), we take the logarithm of both sides:
\[ \log\left(\left( \frac{2}{3} \right)^n\right) < \log\left(\frac{1}{6}\right) \]
- This simplifies to:
\[ n \cdot \log\left(\frac{2}{3}\right) < \log\left(\frac{1}{6}\right) \]

5. **Calculate the Logarithms**:
- Since \( \log\left(\frac{2}{3}\right) \) is negative, we can divide by it, which will reverse the inequality:
\[ n > \frac{\log\left(\frac{1}{6}\right)}{\log\left(\frac{2}{3}\right)} \]

6. **Evaluate the Right-Hand Side**:
- We can calculate \( \log\left(\frac{1}{6}\right) \) and \( \log\left(\frac{2}{3}\right) \):
- \( \log\left(\frac{1}{6}\right) = \log(1) - \log(6) = 0 - \log(6) = -\log(6) \)
- \( \log\left(\frac{2}{3}\right) = \log(2) - \log(3) \)
- Thus:
\[ n > \frac{-\log(6)}{\log(2) - \log(3)} \]

7. **Approximate Values**:
- Using approximate values \( \log(2) \approx 0.301 \) and \( \log(3) \approx 0.477 \):
- \( \log(6) = \log(2) + \log(3) \approx 0.301 + 0.477 = 0.778 \)
- Therefore:
\[ n > \frac{0.778}{0.301 - 0.477} = \frac{0.778}{-0.176} \approx -4.42 \]
- Since \( n \) must be a positive integer, we need to check integer values.

8. **Check Integer Values**:
- Testing \( n = 5 \):
\[ \left( \frac{2}{3} \right)^5 = \frac{32}{243} \approx 0.1317 < \frac{1}{6} \approx 0.1667 \]
- Testing \( n = 4 \):
\[ \left( \frac{2}{3} \right)^4 = \frac{16}{81} \approx 0.1975 > \frac{1}{6} \]
- Therefore, the minimum \( n \) that satisfies the condition is \( n = 5 \).

### Final Answer:
The minimum number of independent shots required is \( \boxed{5} \).