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# PS-Probability -school club

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Senior Manager
Joined: 05 Jun 2008
Posts: 304

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07 Dec 2008, 02:22
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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
Manager
Joined: 14 Oct 2008
Posts: 160

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07 Dec 2008, 08:06
1
KUDOS
Is it C ?

Among 1st Year : 16 Beer, 16 Mixed , Both = 8, Neither = 0 Total = 40
Among 2nd Year : 18 Beer, 18 Mixed , Both = 12, Neither = 12 Total = 60

Total Beer =34 , Mixed =34 , Both = 20

Hence probability = total drinking mixed / total drinking beer
= 20 / 34 = 10 /17

Whats the QA ?
Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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07 Dec 2008, 08:10
Ans is C.

Same explanation that gamecode provided.
Intern
Joined: 26 Sep 2008
Posts: 19

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07 Dec 2008, 20:17
I agree with gameCode that:
Among 1st Year : 16 Beer, 16 Mixed , Both = 8, Neither = 0 Total = 40
Among 2nd Year : 18 Beer, 18 Mixed , Both = 12, Neither = 12 Total = 60

Total Beer =34 , Mixed =34 , Both = 20
but I also disagree that :
probability = total drinking mixed / drinking beer
= 20 / 34 = 10 /17
I think the total drinking beer is 34+20 = 54
then probability = 20/54=10/27 ( not see in the answers)

What happen if the question is :
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, 10% are drinking beer, 40% are drinking mixed drinks, and 50% are drinking both. Of the second year students, 10% are drinking beer, 30% are drinking mixed drinks, and 40% 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?

then we have
Among 1st Year : 4 Beer, 16 Mixed , Both = 20, Neither = 0 Total = 40
Among 2nd Year : 6 Beer, 18 Mixed , Both = 24, Neither = 12 Total = 60

Total Beer =10 , Mixed =34 , Both = 44
if we count probability = total drinking mixed / drinking beer then
probability = 44 / 10 = 4.4 > 1 ( It's imposible)
Re: PS-Probability -school club   [#permalink] 07 Dec 2008, 20:17
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