Probability Question

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).

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

Thanks a lot, i was really confused with what was the part i was looking for and the total #