Need clarification

[callmemo]

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

OA: C. IMO, the answer should have been 10/27. thoughts??

[4meonly]

I got 20/100 = 1/5


I think Q asks about
(number of students drinking both) / (total number of students)
let 100 be total # of students
1st year students:
40 - total, 20%*40=8 drinking both

2nd year:
60 total, 20%*60=12 drinking both

(8+12)/100 = 20/100 = 1/5

Where is my mistake?

[parallel_chase]

First year student = 40
Beer drinkers = 16
Mixed drinkers = 16
Both = 8

Second year students = 60
Beer drinkers = 18
Mixed drinkers = 18
Both = 12

Now the question says "If the student is drinking beer, what is the probability that he or she is also drinking mixed drinks?"

both (beer + mixed) = 8 + 12 = 20

only beer + both = 16+18+8+12 = 54

probability = 20/54 = 10/27

I am not sure about the framing of the question. But this is the only way you can get 10/27 as the answer. Whats the source ?

[callmemo]

Delta Course, Daily GMAT Practice Question

https://www.deltacourse.com/questions/512341ZF.asp

[4meonly]

strange Q
From the posted link:
(16% + 18%) or 34% of the group is drinking beer.
(8% + 12%) or 20% of the group is drinking both beer and mixed drinks.
If a student is chosen at random is drinking beer, the probability that they are also drinking mixed drinks is (20/34) or 10/17.

(drinking beer)/(drinking both beer and mixed drinks)
It should be
(drinking beer)/(drinking beer+drinking both beer and mixed drinks)

[mental]

at first try I had landed on 10/27

but reading again (as the answer was not in choice) i probably found my mistake in interpretation

Ques says 40% Beer, 40% mixed, nd 20% both
it is not saying "ONLY"

lets say total B school students=100

40 1st yr...............................................60 2nd yr
16 beer.................................................18 beer
16 mixed...............................................18 mixed
among them(32) 8 both..........................among them(36) 12 both

so if the student was drinking beer = 16 + 18 = 34
both........8 + 12 = 20

prob: 20/34 = 10/17
........................................................................................................

[ab78]

Just to help with more clarification this is a conditional probability problem.

P(Beer)=P(Beer intersection Mixed Drinks) /P(Mixed Drinks) essentially you can write it as

P(Beer) = # of students drinking both /# of students drinking mixed drinks

why?? because total # of students is the same in both case and hence it cancels out.

[kris610]

I go with 10/17.

If you use numbers 40 and 60 for 1st and 2nd year students you will see:

34 drink beer and 20 drink both.

Since, you know that the selected person drinks beer, 34 will be the sample space and 20/34 i.e. 10/17 will be the required probability.