Question:
The probability of a man hitting a target is \(\frac{2}{5}\). He fires at the target \(k\) times (\(k\), a given number). Then the minimum \(k\), so that the probability of hitting the target at least once is more than \(\frac{7}{10}\), is :
The probability of a man hitting a target is \(\frac{2}{5}\). He fires at the target \(k\) times (\(k\), a given number). Then the minimum \(k\), so that the probability of hitting the target at least once is more than \(\frac{7}{10}\), is :
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Use the complement: P(at least one hit) = 1 - P(no hit in all k shots).
Updated On: Oct 31, 2026
- 3
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The Correct Option is A
Solution and Explanation
\(P(\text{hit}) = \frac25\), \(P(\text{miss}) = \frac35\).
\(P(\text{at least one hit in } k \text{ shots}) = 1 - \left(\frac35\right)^k \gt \frac{7}{10}\) \(\Rightarrow\) \(\left(\frac35\right)^k \lt \frac{3}{10} = 0.3\).
\(k=1\): 0.6; \(k=2\): 0.36; \(k=3\): 0.216 \(\lt 0.3\).
The smallest such \(k\) is 3.
Answer: (1) 3
\(P(\text{at least one hit in } k \text{ shots}) = 1 - \left(\frac35\right)^k \gt \frac{7}{10}\) \(\Rightarrow\) \(\left(\frac35\right)^k \lt \frac{3}{10} = 0.3\).
\(k=1\): 0.6; \(k=2\): 0.36; \(k=3\): 0.216 \(\lt 0.3\).
The smallest such \(k\) is 3.
Answer: (1) 3
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