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Suppose that a bagel shop sells eight different kinds of bagels and we want to choose a dozen bagels. Suppose the shop packs a dozen bagels for you at random. What is the probability that it contains at least two egg bagels and no more than six salty bagels?

What I have done so far:

Total = ${12 + 8 − 1 \choose 12} = {19 \choose 12}$

At least two egg bagels = ${10 + 8 − 1 \choose 10}/{19 \choose 12}$

No more than six salty bagels = $1 − {5 + 8 − 1 \choose 5}/{19 \choose 12}$

Based on these how I can calculate "at least two egg bagels and no more than six salty bagels" without over counting ?

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seriously ,No one ! –  Hooman Dec 14 '12 at 3:21
Your attempted solution underlines the fact that "randomly" is not a complete specification because it doesn't specify the distribution. Even assuming that you mean "uniformly randomly", it leaves open the question what should be uniformly distributed. Your attempted solution appears to assume that the combination of kinds should be uniformly randomly picked from all combinations of kinds, which is certainly not what I would have expected from "packs at random", which to me sounds more like choosing each individual bagel uniformly randomly from the eight kinds. –  joriki Dec 14 '12 at 4:31

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