probability q

Sadowski wrote:Say that 100 students are at the party.

40 - 1st year
16 - beer
16 - mixed drinks
8 - both

60 - 2nd year
18 - beer
18 - mixed drinks
12 - both

Total beer drinkers = 16+18=34
Total both = 12+8=20

20/34 = 10/17

Is C the correct answer?

Sadowski wrote:If you added the beer drinkers and the "both" drinkers in the denominator, you would be adding some people twice.

Based on the question, you first need to find out how many total possible people could be drinking beer. Because the "both" drinkers are already counted in the "beer" and "mixed drink" categories, we know that the sum of the 40% of first years and 30% of second years is the total amount of beer drinkers at the party.

That may not be perfectly clear, but I hope it helps.

gabriel wrote:

Sadowski wrote:If you added the beer drinkers and the "both" drinkers in the denominator, you would be adding some people twice.

Based on the question, you first need to find out how many total possible people could be drinking beer. Because the "both" drinkers are already counted in the "beer" and "mixed drink" categories, we know that the sum of the 40% of first years and 30% of second years is the total amount of beer drinkers at the party.

That may not be perfectly clear, but I hope it helps.

This actually is not entirely true .. if u notice, for the first year students the addition of the different categories ( that is people having beer, mixed drinks, and both ) comes exactly to 100% .. so we cant conclude over here that the people who drink both have already been included in the people who drink beer .. since this part is not mentioned by the author of the question there is a certain degree of ambiguity in the question .. anyway i dont think we will see such questions on the GMAT ...