A jar contains 6 chocolate chip cookies and 9 peanut butter cookies. Richard grabs 3 cookies at random to pack in his lunch. What is the probability that he drew 2 chocolate chip cookies and 1 peanut butter cookie?

Question

kirbykirby · Answer

If you are familiar with the hypergeometric distribution, it's basically this. If not, step by step:

Since the order of what you pick doesn't matter... then the number of ways to choose 2 chocolate cookies from 6 is $$ {6 \choose 2}$$
To get 1 peanut butter cookie, this is $$ {9 \choose 1} $$
So, the number of ways to get 2 chocolate and 1 peanut is:
$$ {6 \choose 2}{9 \choose 1}$$. Now, the sample space is the total number of cookies you pick out of 15, and there are $$ {15 \choose 3}$$ ways to do so. So the probability is:
$$ \large \frac{{6 \choose 2}{9 \choose 1}}{{15 \choose 3}}$$

anonymous · Answer

So, it is $$\frac{ 27 }{ 91 }$$

kirbykirby · Answer

Another way to think about it:
Pretend you are picking Chocolate, CHocolate, Peanut consecutively... 
1st draw: P(choco) = 6/15
2nd draw: P(choco) = 5/14, since there is one chocolate missing now
3rd draw: P(peanut) = 9/13, again since another chocolate is missing
So P(choco, choco, peanut) = 6/15*5/14*9/13 = 9/91

But this is the arrangement CCP. In fact, we do not care about the order... you chould have also picked:
CPC
or PCC

Since there are 3 more combinations possible, you multiply 9/91 by 3 
(You can think of it like 
P(2 choco and 1 peanut)=Prob(CCP) or Prob(CPC) or Prob(PCC) = 3*Prob(CCP)

kirbykirby · Answer

yes

anonymous · Answer

Awesome, thanks. In the first steps, it reminded me of combinations. (6C2*9C1)/15C3

kirbykirby · Answer

indeed =] yw