SOLUTION: 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 probabili

Algebra ->  Permutations -> SOLUTION: 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 probabili      Log On


   



Question 1122550: 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

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Number of ways of choosing 3 of the 9 chocolates: 9C3 (9 choose 3).

Number of ways of choosing 2 of the 3 with coconut (and 1 of the other 7): (3C2)*(7C1).

The probability is the number of ways of getting the desired combination, divided by the total number of possible combinations.

Do the arithmetic and simplify the resulting fraction.

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
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  C%5B9%5D%5E3 = %289%2A8%2A7%29%2F%281%2A2%2A3%29 = 84.

    It is the number of all elements in the "space events".



3.  Spider-man can choose 2 of 3 coconut chocolates by  C%5B3%5D%5E2 = 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  18%2F84 = 3%2F14.