# Probability of hiting the target

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?

• $1-0.8^{10}$ means hitting target at least once. You want $P(H_2|H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)}$ – hjpotter92 Sep 15 '15 at 12:09

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