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Consider. You throw 3 baseballs at a target. Each baseball has the probability (independent) p= 3/5 of a hit. What's the probability that all three hit? If it is such that a target will be knocked over if 2 or more hit, what is the probability that target gets knocked over.

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1 Answer 1

Since throws are independant, the probability of all $3$ hitting is $$ \left(\frac{3}{5}\right)^3=\frac{27}{125}=0.216 $$ Now for the second part, let's assume that we only need $2$ hits in order for the targer to fall, we have to choose where those $2$ hits occur since it will affect the probability.

Say we hit on our first $2$ throws. As above, this occurs with probability $\left(\frac{3}{5}\right)^2$. We won't throw a third ball because there is no target. Now we could also have the sequence [hit,miss,hit] or [miss, hit,hit]. Both of these occur with probability $\frac{2}{5}\left(\frac{3}{5}\right)^2$. So total, you get $$ \left(\frac{3}{5}\right)^2+2\times \frac{2}{5}\left(\frac{3}{5}\right)^2=\frac{81}{125}=0.648 $$

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Although not tagged as such this question looks like a homework problem. Even though this is a simple problem to answer shouldn't we be given hints about getting the solution rather than supply a complete solution as you have done here? – Michael Chernick Sep 30 '12 at 13:41
yeah maybe it is homework, I'll be more careful next time ;) – Jean-Sébastien Sep 30 '12 at 14:08

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