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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



        The solution by the tutor @greenestamps was not exactly correct,


        so I came to fix and finalize it in accurate way.



<pre>
1.  Notice that 4 + 3 + 2 = 9 chocolates.



2.  The number of all possible triples of chocolates is  {{{C[9]^3}}} = {{{(9*8*7)/(1*2*3)}}} = 84.

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



3.  Spider-man can choose 2 of 3 coconut chocolates by  {{{C[3]^2}}} = 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  {{{18/84}}} = {{{3/14}}}.
</pre>