BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability Question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 75
Joined: Fri Jan 16, 2009 2:00 pm
Thanked: 1 times
GMAT Score:550

Probability Question

by jfranco23 » Thu Mar 05, 2009 1:50 pm
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
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Probability Question

by Stuart@KaplanGMAT » Thu Mar 05, 2009 2:12 pm
jfranco23 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
Great question for picking numbers!

let's say we have 100 total students. Therefore, we have:

40 first year, of whom:
40% = 16 drink beer
40% = 16 drink mixed drinks
20% = 8 drink both

60 second year, of whom:

30% = 18 drink beer
30% = 18 drink mixed drinks
20% = 12 drink both

Q: if a student is drinking beer, what's the probability that s/he is also drinking mixed drinks?

Prob = # of desired/total #

Total # of both = 8+12 = 20
Total # of beer = 16+18 = 34

Therefore, our answer is 20/34 = 10/17... choose (C).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 160
Joined: Fri May 30, 2008 7:10 pm
Thanked: 10 times
GMAT Score:600

by dendude » Thu Mar 05, 2009 2:24 pm
This needs to be solved by Conditional Probability, which deals with two events occurring when there's a dependenency

P(B|A) = P(A and B)/P(A)
Here P(A and B) is the probability of someone drinking both drinks.
P(A) is the probability of someone drinking Beer

Based on the info in the q,
Ist Year IInd Year
40 60
Beer 16(40%) 18(30%)
Mixed 16(40%) 18(30%)
Both 8(20%) 12(20%)

Using this,
P(A and B) = (8+12)/100 = 20/100
P(B) = (16+18)/100= 34/100

P(B|A) = 20/34 = 10/17
Top
Senior | Next Rank: 100 Posts
Posts: 75
Joined: Fri Jan 16, 2009 2:00 pm
Thanked: 1 times
GMAT Score:550

by jfranco23 » Thu Mar 05, 2009 2:24 pm
Thanks a lot, i was really confused with what was the part i was looking for and the total #
Top
Post new topic Post Reply
• Page 1 of 1