# 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:

Favorable outcomesPossible outcomes\frac{Favorable\ outcomes}{Possible\ outcomes}
Since I only care about choosing three cookies, I think the possible outcomes are (103)\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 (43)=4\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}.\frac{1+1+\binom{4}{3}}{\binom{10}{3}}=\frac{6}{120}=\frac{1}{20}.

But I’m not sure if I’m correct.

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I’d solve it, but you need to show some initial thoughts of yours…
– Parcly Taxel
2 days ago

I just edited. Sorry, I forgot to post what I tried.
– Rigo
2 days ago

1

Well done! Your solution is great.
– Thanassis
2 days ago

1

– Parcly Taxel
2 days ago

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1

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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}\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}\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}\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}

Sum it up and you get \frac{36}{720} = \frac{1}{20}\frac{36}{720} = \frac{1}{20}

Thank you! I have a question, shouldn’t P( 3 * vanilla)P( 3 * vanilla) be \frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}\frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}? And the same with the other three equalities?
– Rigo
2 days ago

Actually 36/72036/720, because the second cookie is selected at random from just 99 cookies remaining in the jar, not 1010, and the third cookie is selected from 88 cookies.
– David K
2 days ago

@Rigo Yes, it’s exactly like you said.
– David K
2 days ago

Yeah ! Your’re right ! Thank you. I was already wondering …
– Jakov Kholodkov
2 days ago