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If the probability of hitting a target is $1/5$, and 10 shots are fired independently, what is the probability of the target being hit at least twice?

Let $X$ be the number of times the target is hit. We want to find $P(X \geq 2)$. So $$P(X \geq 2) = 1-P(X < 2)$$

$$ = 1- \binom{10}{0} (0.2)^{0} (0.8)^{10}-\binom{10}{1} (0.2)^{1}(0.8)^{9}$$

Would that be correct?

Thanks

  • 2
    Yes, that is correct, and I am pleased that you figured out that it is easier to calculate the complementary probability instead of adding up $P\{X = 2\} + P\{X = 3\} + \cdots + P\{X = 10\}$2011-12-29

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