Hi ziyuenlau,

I believe there's an error/typo in this question (the correct answer doesn't appear among the 5 choices). The math behind this question can be approached in a couple of different ways.

We have 4 cans each of: cola, root beer and ginger ale - which gives us a total of 12 cans. We're asked for the probability of pulling 2 cans and getting 1 cola and one root beer.

1st can = cola OR root beer = 8/12

Once we pull that first can, there will then be 11 cans remaining and we'll need to pull the OTHER type (re: if we pulled cola 1st, then we would need to pull root beer 2nd - and vice versa).

2nd CAT = the other type = 4/11

Thus, the probability = (8/12)(4/11) = 32/132 = 8/33

GMAT assassins aren't born, they're made,
Rich

ziyuenlau wrote:Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 2/11

D. 3/11

E. 1/3

I agree with Rich - the correct answer is not listed. But here is my solution:

Let's first determine the number of ways Jerome can pull 1 can of root beer and 1 can of cola from the fridge. The number of ways to pull the cola is 4C1 = 4 and the number of ways to pull the root beer is 4C1 = 4.

The total number of ways to pull 2 cans of any beverage from the fridge is 12C2 = (12 x 11)/2! = 66 ways.

Thus, the probability is (4 x 4)/66 = 16/66 = 8/33.