The probabilty of hitting a target is 1/3. The least number of times to fire so that the probability of hitting the larget atleast once is more than `90%` is

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the problem, we need to determine the least number of times to fire at a target such that the probability of hitting it at least once is more than 90%.

### Step-by-Step Solution:

1. **Identify the Probability of Hitting and Not Hitting the Target**:
- The probability of hitting the target (P(hit)) is given as \( \frac{1}{3} \).
- Therefore, the probability of not hitting the target (P(not hit)) is:
\[ P(\text{not hit}) = 1 - P(\text{hit}) = 1 - \frac{1}{3} = \frac{2}{3} \]

2. **Set Up the Probability Condition**:
- We want the probability of hitting the target at least once in \( n \) tries to be greater than 90%. This can be expressed as:
\[ P(\text{at least one hit}) > 0.9 \]
- The probability of hitting at least once can be calculated using the complement rule:
\[ P(\text{at least one hit}) = 1 - P(\text{not hit in } n \text{ tries}) = 1 - \left( P(\text{not hit}) \right)^n \]

3. **Express the Condition Mathematically**:
- Substituting the values we have:
\[ 1 - \left( \frac{2}{3} \right)^n > 0.9 \]
- Rearranging this gives:
\[ \left( \frac{2}{3} \right)^n < 0.1 \]

4. **Calculate Values of \( n \)**:
- We need to find the smallest integer \( n \) such that \( \left( \frac{2}{3} \right)^n < 0.1 \).
- We can calculate \( \left( \frac{2}{3} \right)^n \) for various values of \( n \):
- For \( n = 1 \):
\[ \left( \frac{2}{3} \right)^1 = \frac{2}{3} \approx 0.6667 \]
- For \( n = 2 \):
\[ \left( \frac{2}{3} \right)^2 = \frac{4}{9} \approx 0.4444 \]
- For \( n = 3 \):
\[ \left( \frac{2}{3} \right)^3 = \frac{8}{27} \approx 0.2963 \]
- For \( n = 4 \):
\[ \left( \frac{2}{3} \right)^4 = \frac{16}{81} \approx 0.1975 \]
- For \( n = 5 \):
\[ \left( \frac{2}{3} \right)^5 = \frac{32}{243} \approx 0.1317 \]
- For \( n = 6 \):
\[ \left( \frac{2}{3} \right)^6 = \frac{64}{729} \approx 0.0878 \]

5. **Determine the Minimum \( n \)**:
- From the calculations, we see that:
- For \( n = 5 \), \( \left( \frac{2}{3} \right)^5 \approx 0.1317 \) (not less than 0.1)
- For \( n = 6 \), \( \left( \frac{2}{3} \right)^6 \approx 0.0878 \) (less than 0.1)
- Therefore, the least number of times to fire so that the probability of hitting the target at least once is more than 90% is \( n = 6 \).

### Final Answer:
The least number of times to fire so that the probability of hitting the target at least once is more than 90% is **6**.