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A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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20 Apr 2010, 08:11

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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?

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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

I didn't understand what the question asked . I tried to calculate but all mismatch. Please provide some explanations.

Let's say total students =100 First year=40 Second year=60

Out of the first year students: 16 drink beer (40% of 40), 16 drink mixed drinks (40% of 40) and out of these 32 students, 8 are drinking both (20% of 40). What are other 8 students drinking, we don't know.

Out of the second year students: 18 drink beer (30% of 60), 18 drink mixed drinks (30% of 60) and out of these 36 students, 12 are drinking both (20% of 60). What are other 24 students drinking, we don't know.

So total beer drinkers are 16+18=34. Out of these students 8+12=20 are drinking both beer and mixed drinks. So probability that student who is drinking beer also is drinking mixed drinks is 20/34=10/17.

Re: A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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20 Apr 2010, 08:55

Bunuel wrote:

ykaiim wrote:

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

I didn't understand what the question asked . I tried to calculate but all mismatch. Please provide some explanations.

Let's say total students =100 First year=40 Second year=60

Out of the first year students: 16 drink beer (40% of 40), 16 drink mixed drinks (40% of 40) and out of these 32 students, 8 are drinking both (20% of 40). What are other 8 students drinking, we don't know. I think you already sum up to 40. Where from this 8 more come?

Out of the second year students: 18 drink beer (30% of 60), 18 drink mixed drinks (30% of 60) and out of these 36 students, 12 are drinking both (20% of 60). What are other 24 students drinking, we don't know. I think it should be 12.

So total beer drinkers are 16+18=34. Out of these students 8+12=20 are drinking both beer and mixed drinks. So probability that student who is drinking beer also is drinking mixed drinks is 20/34=10/17.

Answer: C.

One more thing, is this question asking for the ratio of above two calculations?
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One more thing, is this question asking for the ratio of above two calculations?

Your calculation is not correct.

Step by step:

Out of the first year students: 16 drink beer (40% of 40), 16 drink mixed drinks (40% of 40) --> 16+16=32. And out of these 32 students, 8 are drinking both (20% of 40).

40-32=8 and we don't know what are they drinking (actually it doesn't matter).

Out of the second year students: 18 drink beer (30% of 60), 18 drink mixed drinks (30% of 60) --> 18+18=36. And out of these 36 students, 12 are drinking both (20% of 60).

60-36=24 and we don't know what are they drinking (actually it doesn't matter).

So total beer drinkers are 16+18=34. Out of these students 8+12=20 are drinking both beer and mixed drinks. So probability that student who is drinking beer (denominator) also is drinking mixed drinks (numerator) is: 20/34=10/17.
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Re: A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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20 Apr 2010, 23:03

ykaiim wrote:

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

I didn't understand what the question asked . I tried to calculate but all mismatch. Please provide some explanations.

hi, probability of an event = favorable outcomes / total outcomes suppose there are 1000 students (1000 taken to simplify the calculations) first year students = 40% of 1000 = 400 second year students = 60% of 1000 = 600

of first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both => the number is 40% of 400, 40% of 400 and 20% of 400 respectively => 160, 160, 80

of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both => the number is 30% of 600, 30% of 600 and 20% of 600 respectively => 180, 180, 120

total (both beer and mixed drinks) drinkers = 80 + 120 = 200 total beer drinkers = 160 + 180 = 340

required probability = 200/340 = 10/17
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Re: A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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20 Apr 2010, 23:39

Bunuel and GurpeetSingh.

Thanks for your analysis. I am clear now whre I was wrong. Both of you provided great insights to this problem.
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A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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10 Jul 2011, 20:43

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?

Re: A business school club, Friends of Foam, is throwing a party at a loca [#permalink]

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10 Jul 2011, 22:58

Alchemist1320 wrote:

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

Suppose there are 100 students

Group A : 40% = 40 students 40% drink beer = 16 40% mixed = 16 20% both = 8

Group B 60% = 60 30% beer= 18 30% mixed = 18 20% both= 12

now we need both ( beer + mixed = both)

probability = total beer drinker = 16+18 =34 and both = 20 thus 20/34 = 10/17 Hence C

gmatclubot

Re: A business school club, Friends of Foam, is throwing a party at a loca
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10 Jul 2011, 22:58

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