| A can hit a target 4 times in 5 shots; |
| B can hit a target 3 times in 4 shots; |
| C can hit a target 2 times in 3 shots; |
| All the three a shot each. What is the probability that two shots are at least hit? |
A) 1/6
B) 3/5
C) 5/6
D) 1/3
Correct Answer: C
Solution :
[c] Probability of no one hitting the target \[=\frac{1}{5\times 4\times 3}=\frac{1}{60}\] Probability of one hitting the target \[=\frac{4+3+2}{60}=\frac{9}{60}\] \[\therefore \] Probability of maximum one hit \[=\frac{1}{60}+\frac{9}{60}=\frac{10}{60}=\frac{1}{6}\] Probability that two shots are hit at least is the required probability\[=1-\frac{1}{6}=\frac{5}{6}\]
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