A business school club, Friends of Foam, is throwing a party at a local bar. Of the business school students at the bar, 40% are first year students and 60% are second year students. Of the first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both. Of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both. A business school student is chosen at random. If the student is drinking beer, what is the probability that he or she is also drinking mixed drinks?
A. 2/5
B. 4/7
C. 10/17
D. 7/24
E. 7/10
source: the delta course
probability q
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- Sadowski
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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?
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?
yep, the correct answer's CSadowski 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?
can you explain why it's 20/34, instead of 20/54? I got 54 from adding total beer drinkers and total both.
thanx for your answer.
- Sadowski
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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.
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
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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 ...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.
- Sadowski
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I agree with you. I had to make an assumption when I started to solve the problem. And I also agree that the GMAT will give "air-tight" problems that don't require extra assumptions to be able to solve the problem.gabriel wrote: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 ...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.
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Hmm, I took it as a probability problem.
I first got the probability of choosing beer only drinkers in the room:
40/100 * 40/100 + 60/100 * 30/100 = 34/100
Then, I got the probability of choosing beer and mixed drinkers in the room:
40/100 * 20/100 + 60/100 * 20/100 = 20/100
Finally, to solve the problem, you have to get the probability of choosing a beer and mixed drinker out of the beer only drinkers:
(20/100) / (34/100) = 20/34 = 10/17.
I first got the probability of choosing beer only drinkers in the room:
40/100 * 40/100 + 60/100 * 30/100 = 34/100
Then, I got the probability of choosing beer and mixed drinkers in the room:
40/100 * 20/100 + 60/100 * 20/100 = 20/100
Finally, to solve the problem, you have to get the probability of choosing a beer and mixed drinker out of the beer only drinkers:
(20/100) / (34/100) = 20/34 = 10/17.