<s>[INST] Olympic problem - Probability A jar contains 3 cookies filled with chocolate, 4 cookies filled with strawberry and 3 cookies filled with vainilla. What is the probability of choosing 3 cookies with the same filling from the jar?
EDIT: I've tried this:
What I want to find is:
$$\frac{Favorable\ outcomes}{Possible\ outcomes}$$
Since I only care about choosing three cookies, I think the possible outcomes are $\binom{10}{3}$.
Now, the favorable outcomes should be: 6.
Only one option to get the three chocolate filled cookies, same with the vanilla filling. I have $\binom{4}{3}=4$ ways to choose three cockies from the strawberry cookies.
Now, the probability I want should be:
$$\frac{1+1+\binom{4}{3}}{\binom{10}{3}}=\frac{6}{120}=\frac{1}{20}.$$
But I'm not sure if I'm correct.
 [/INST] Look at the problem as a choice tree and ask yourself: how many branches do I have that satisfy the conditions above?
In this case you have a tree with depth three (3). How many branch ends are there to satisfy your condition, and what is the probability of getting to each of them? 
P(3 vanilla) = $\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}$
P(3 straberry) = $\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{24}{720}$
P(3 chocolate) = $\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}$
Sum it up and you get $\frac{36}{720} = \frac{1}{20}$
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