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 the target at least once is greater than \( \frac{5}{6} \).

### Step-by-step Solution:

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

2. **Let \( n \) be the number of shots:**
- We want to find the minimum \( n \) such that the probability of hitting the target at least once is greater than \( \frac{5}{6} \).

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

4. **Set Up the Inequality:**
- We need this probability to be greater than \( \frac{5}{6} \):
\[ 1 - \left(\frac{2}{3}\right)^n > \frac{5}{6} \]

5. **Rearranging the Inequality:**
- Subtract 1 from both sides:
\[ -\left(\frac{2}{3}\right)^n > -\frac{1}{6} \]
- Multiply both sides by -1 (which reverses the inequality):
\[ \left(\frac{2}{3}\right)^n < \frac{1}{6} \]

6. **Taking Logarithms:**
- Taking logarithm on both sides:
\[ \log\left(\left(\frac{2}{3}\right)^n\right) < \log\left(\frac{1}{6}\right) \]
- This simplifies to:
\[ n \log\left(\frac{2}{3}\right) < \log\left(\frac{1}{6}\right) \]

7. **Solving for \( n \):**
- Since \( \log\left(\frac{2}{3}\right) \) is negative, we can divide both sides by it (which will flip the inequality):
\[ n > \frac{\log\left(\frac{1}{6}\right)}{\log\left(\frac{2}{3}\right)} \]

8. **Calculating the Values:**
- Calculate \( \log\left(\frac{1}{6}\right) \) and \( \log\left(\frac{2}{3}\right) \):
- \( \log\left(\frac{1}{6}\right) = \log(1) - \log(6) = 0 - \log(6) = -\log(6) \)
- \( \log\left(\frac{2}{3}\right) = \log(2) - \log(3) \)

9. **Finding the Minimum \( n \):**
- Using approximate values:
- \( \log(2) \approx 0.301 \)
- \( \log(3) \approx 0.477 \)
- \( \log(6) = \log(2) + \log(3) \approx 0.301 + 0.477 = 0.778 \)
- Therefore:
\[ n > \frac{-0.778}{0.301 - 0.477} = \frac{-0.778}{-0.176} \approx 4.42 \]
- Since \( n \) must be a whole number, we round up to the next whole number:
\[ n = 5 \]

### Conclusion:
The minimum number of independent shots required by the shooter so that the probability of hitting the target at least once is greater than \( \frac{5}{6} \) is **5**.