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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 22 Apr 2005, 20:29
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


Delta's solution & explanation (SPOILER): http://www.deltacourse.com/questions/512341ZF.asp

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 [#permalink] New post 22 Apr 2005, 21:47
C for me...looks okay...lemme see if i can explain this...

Suppose total students 100
First Year(FY) = 40, Second Year (SY) = 60
FY drinking Beer = 40% of 40 = 16
FY drinking Mixed = 40% of 40 = 16
FY drinking Both = 20% of 40 = 8
Therefor, FY drinking None = 40 - (16 + 16 - 8) = 16

SY drinking Beer = 30% of 60 = 18
SY drinking Mixed = 30% of 60 = 18
SY drinking Both = 20% of 60 = 12
Therefore SY drinking none = 36

Now, the student chosen is drinking beer...the total number of students drinking beer = 16 + 18 = 34
Total number of students drinkng both = 8 + 12 = 20 (because our favourable case will be when the student is also from these two groups)

Therefore, the probability that the student is also drinking a mixed drink = 20/34 = 10/17

HTH

cheers!
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 [#permalink] New post 23 Apr 2005, 08:26
makes sense... thanks for the explanation!
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Conditional probability [#permalink] New post 24 Apr 2005, 06:52
P(AnB) = P(A/B)*P(B)

In words,

P(Both A and B occuring) = P(A occuring given B has already occured) * P(B occuring)

In our case A = mixed drink and B = beer.

If we start with 100 then 40 Ist year ( 16 beer , 16 mixed and 8 both)

60 2nd year (18 beer, 18 mixed and 12 both)

Let's find P(B) = P(Beer) = P(1st year)*P(Beer in 1st year) + P(2nd year)*P(Beer in 2nd year)

= 2/5*2/5 + 3/5*3/10 = 17/50.

P(AnB) = P(Both) = (8+12)/(100) = 1/5

So From the equation,

1/5 = P(AIB)*17/50

=> P(AIB) = 10/17

C
Conditional probability   [#permalink] 24 Apr 2005, 06:52
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