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Among $t = 60$ lottery tickets there are $w = 20$ prizes. We buy $b = 6$. What is the probability that $g$ tickets will win, with $g=2$? Generalize this to arbitrary numbers $t,w, b, g$.

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I think it is 5/7 but I'am not sure???? – XTTTX Dec 28 '12 at 17:33
How did you go about computing this number? – Eckhard Dec 28 '12 at 17:36
up vote 1 down vote accepted

There are $\dbinom{60}{6}$ ways to choose $6$ tickets, all equally likely.

There are $\dbinom{20}{2}\dbinom{40}{4}$ ways to choose $2$ winners and therefore $4$ losers.


The same idea works in general.

Another way: Imagine that we buy and check the tickets one at a time. We can get $2$ winners and $4$ losers in various ways. For example, consider the sequence WWLLLL. This has probability $$\frac{20}{60}\frac{19}{59}\frac{40}{58}\frac{39}{57}\frac{38}{56}\frac{37}{55}.\tag{$1$}$$ But we can win precisely twice in various other ways, like WLWLLL. Compute the probability of this. We will get the same result as for WWLLLL. The same holds for every sequence that has exactly $2$ W and $4$ L. There are $\dbinom{6}{2}$ such sequences. So for our probability, multiply the result of $(1)$ by $\dbinom{6}{2}$.

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