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

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Director
Joined: 05 Jan 2005
Posts: 561
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A business school club, Friends of Foam, is throwing a party [#permalink]  11 Jul 2005, 23:21
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
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: 11 Jan 2005
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what about the 20% of the second year left ? what do they drink?
Senior Manager
Joined: 17 Apr 2005
Posts: 375
Location: India
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Got 10/27. No choice corresponds to this.

HMTG.
Intern
Joined: 27 Jun 2005
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Kudos [?]: 2 [0], given: 0

Assume there are 100 students total.
40 First years - breaks down to:
16 first year beer drinkers
16 first year mixed drinkers
8 first year both drinkers

60 Second years:
18 beer
18 mixed
12 both
12 drink nothing?

So there are 16+8=24 first year beer/both drinkers and
18+12=30 second year beer/both drinkers
24+30=54 out of 100 people drinking beer.
4/7 gives 57 people - this is the closest answer. I choose 4/7 or none of the above, heh.
Manager
Joined: 28 Jun 2005
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My interpretation is as follows

40 First years
16 beer drinkers
16 mixed drinkers
8 both drinkers

60 Second years:
18 beer
18 mixed
12 both

Number of beer drinkers 16+18 = 34
number of both drikers = 8+12 = 20

prob = 20/34 = 10/17

Is this correct?
Senior Manager
Joined: 29 Nov 2004
Posts: 486
Location: Chicago
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gotoknow3 wrote:
My interpretation is as follows

40 First years
16 beer drinkers
16 mixed drinkers
8 both drinkers

60 Second years:
18 beer
18 mixed
12 both

Number of beer drinkers 16+18 = 34
number of both drikers = 8+12 = 20

prob = 20/34 = 10/17

Is this correct?

You are finding the ration between people who drink both drinks and people who drink beer, that does not help, it should be 20/54....

I get 10/27 too..
_________________

Fear Mediocrity, Respect Ignorance

Manager
Joined: 06 Aug 2005
Posts: 197
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Kudos [?]: 5 [0], given: 0

gotoknow is exactly right.

Draw the Venn diagram at the last step.

If you assume 100 students.
For year 1, you have 16 in B, 16 in M and 8 in the intersection.
For year 2, you have 18 in B, 18 in M and 12 in the intersection.
In total, you have 34 in B, 34 in M and 20 in the intersection.

You are only talking about 48 students !

If I= B intersection M
14 in B-I, 14 in M-I and 20 in I

Question asks for I / B = 20/34 = 10/17
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