Can you please site the source of this question?

Here is my take:
RAS = Room A – Seniors = 1/6
RAR = Room A - rest of the class = 5/6
RBS = Room B – Seniors = 5/7
RBR = Rest of the class in room B = 2/7
RCS = Room C – Seniors = 1/2
RCR = Room C - ½

1. (RAS)(RBS)(RCR) = (1/6)(5/7)(1/2) = 5/84
2. (RAS)(RBR)(RCS) = (1/6)(2/7)(1/2) = 2/84
3. (RAR)(RBS)(RCS) = (5/6)(5/7)(1/2) = 25/84

Add them:
(5 + 2 + 25)1/84 = 32/84

Answer = E

ray wrote:Can you please site the source of this question?

Here is my take:
RAS = Room A – Seniors = 1/6
RAR = Room A - rest of the class = 5/6
RBS = Room B – Seniors = 5/7
RBR = Rest of the class in room B = 2/7
RCS = Room C – Seniors = 1/2
RCR = Room C - ½

1. (RAS)(RBS)(RCR) = (1/6)(5/7)(1/2) = 5/84
2. (RAS)(RBR)(RCS) = (1/6)(2/7)(1/2) = 2/84
3. (RAR)(RBS)(RCS) = (5/6)(5/7)(1/2) = 25/84

Add them:
(5 + 2 + 25)1/84 = 32/84

Answer = E

Take 32/84 one step further and divide the top and bottom by 4 and you get 8/21, answer D.

Thanks, guys. I had the same line of reasoning to get 8/21, D. My only difference was I factored out the 1/2 since it appears in all three parts of the expression. That allows you to reach the answer faster because you stay in 42nds rather than 84ths.

The question source is Kaplan GMAT Course Book 07-08 Edition, page 279, question 46. I modified it slightly by adding Room C. The original question gives the same probabilities for rooms A and B, and asks: if one student is chosen at random from each room (A and B), what is the probability that exactly one of the students chosen is a senior? The correct answer is 9/14.