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A business school club, Friends of Foam, is throwing a party

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A business school club, Friends of Foam, is throwing a party [#permalink] New post 04 Aug 2008, 09:45
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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
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Re: PS Beer [#permalink] New post 04 Aug 2008, 09:47
I think the answer is "It depends on who is paying for it! Someone else buys it 100% chance."
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Re: PS Beer [#permalink] New post 04 Aug 2008, 09:55
jallenmorris wrote:
I think the answer is "It depends on who is paying for it! Someone else buys it 100% chance."


:lol: totally agree !!! but its not OA :wink:
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Re: PS Beer [#permalink] New post 04 Aug 2008, 10:06
Well, GMAC can't get everything correct!

durgesh79 wrote:
jallenmorris wrote:
I think the answer is "It depends on who is paying for it! Someone else buys it 100% chance."


:lol: totally agree !!! but its not OA :wink:

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Re: PS Beer [#permalink] New post 04 Aug 2008, 11:02
10/17

say total no of students =100

no of first yrs = 40
no of second yrs = 60

out of these 40 first yrs 16 drink beer, 16 drink mix and 8 drink both.
similarly out of 60 second yrs, 18 drink beer, 18 drink mix and 12 drink both.

now 34 drink beer and 20 drink both.

so our probability is 20/34 or 10/17
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Re: PS Beer [#permalink] New post 04 Aug 2008, 14:50
well said moriss :lol:

i will go with C too ..
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Re: PS Beer [#permalink] New post 04 Aug 2008, 17:48
This one is a conditional probability.. P(A|B) = P(A int B) / P(B) = prob A given B

here we require P[ (drinking mixed)|(drinking beer)] = P(drinking both)/Probability(drink beer)

let us say we have 100

40 first years - 16 beer 16 mix drink 8 both
60 sec years - 18 beer 18 mix drink 12 both 12 nothing

total beers 34
total mix drinks 34
total boths 20
total nothings 12

now our prob = prob(both)/prob(beer) = (20/100) / (34/100) = 10/17
Re: PS Beer   [#permalink] 04 Aug 2008, 17:48
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