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A can hit a target 4 times in 5 shots; $\mathrm{B}$ can hit a target 3 times in 4 shots:
$\mathrm{C}$ can hit a target 2 times in 3 shots All the three fire a shot each. What is the probability that two shots are at least hit?
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$\mathrm{C}$ can hit a target 2 times in 3 shots All the three fire a shot each. What is the probability that two shots are at least hit?
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Verified Answer
The correct answer is:
$5 / 6$
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 $=1-\frac{1}{6}=\frac{5}{6}$
$=\frac{1}{5 \times 4 \times 3}=\frac{1}{60}$
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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