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A box contains nine chocolates -- four with pecans, three with coconut, and two with raspberry creme.
Spider-Man eats three of the chocolates, chosen randomly. What's the probability that he eats (exactly) two with coconut?
Enter your answer as a fraction in lowest terms
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The solution by the tutor @greenestamps was not exactly correct,
so I came to fix and finalize it in accurate way.
1. Notice that 4 + 3 + 2 = 9 chocolates.
2. The number of all possible triples of chocolates is = = 84.
It is the number of all elements in the "space events".
3. Spider-man can choose 2 of 3 coconut chocolates by = 3 ways, and then he can add any of remaining NON-COCONUT chocolates by 4+2 = 6 ways.
It gives him 3*6 = 18 ways to get 3 chocolates under the condition imposed by the problem.
4. So the answer to the problem question is = .