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If the probability of hitting a target is 1/5, and ten shots are fired independently, what is the probability that the target is hit at least twice? What is the conditional probability that the target is hit at least twice, given that it is hit at least once?

I am interested in the second part(conditional probability) The answer, that book give is: $$(1-{0.8}^{10} - 2*(0.8)^9)/(1 - (0.8)^{10})$$ But i don't understand why we should put in numerator $0.8^{10}$? It is given that it is hit at least once, hence ${0.8}^{10}$(which means no shots hit the target) should be excluded? Am I right?

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    $\begingroup$ $1-0.8^{10}$ means hitting target at least once. You want $P(H_2|H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)}$ $\endgroup$ – hjpotter92 Sep 15 '15 at 12:09
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It IS being excluded, that's why there's a minus sign.

The numerator is just P(at least two hits), which equals:

1 - P(0 hits) - P(1 hit) = $1 - (.8)^{10} - 10*(.2)*(.8)^9$

The denominator is P(at least one hit), which is just

1 - P(0 hits) = $1 - (.8)^{10}$

Did that clear it all up?

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