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

Your calculation is not correct.

Step by step:

Out of the first year students: 16 drink beer (40% of 40), 16 drink mixed drinks (40% of 40) --> 16+16=32. And out of these 32 students, 8 are drinking both (20% of 40).

40-32=8 and we don't know what are they drinking (actually it doesn't matter).

Out of the second year students: 18 drink beer (30% of 60), 18 drink mixed drinks (30% of 60) --> 18+18=36. And out of these 36 students, 12 are drinking both (20% of 60).

60-36=24 and we don't know what are they drinking (actually it doesn't matter).

So total beer drinkers are 16+18=34. Out of these students 8+12=20 are drinking both beer and mixed drinks. So probability that student who is drinking beer (denominator) also is drinking mixed drinks (numerator) is: 20/34=10/17.

Probability = required outcomes/total outcomes

required outcomes = 20% of 40% of total + 20% of 60% of total

= 20% of (40% * x) + 20% of (60%*x) = 20% of x = .2x

total outcomes = total of those who drink beer

= 40% of 40% of x + 30% of 60% of x

= .16x + .18x = .34x

so probability = .2x/.34x = .1/.17 = 10/17

hence C

hi,
probability of an event = favorable outcomes / total outcomes
suppose there are 1000 students (1000 taken to simplify the calculations)
first year students = 40% of 1000 = 400
second year students = 60% of 1000 = 600

of first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both
=> the number is 40% of 400, 40% of 400 and 20% of 400 respectively
=> 160, 160, 80

of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both
=> the number is 30% of 600, 30% of 600 and 20% of 600 respectively
=> 180, 180, 120

total (both beer and mixed drinks) drinkers = 80 + 120 = 200
total beer drinkers = 160 + 180 = 340

required probability = 200/340 = 10/17

Bunuel and GurpeetSingh.

Thanks for your analysis. I am clear now whre I was wrong. Both of you provided great insights to this problem.

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

Suppose there are 100 students

Group A : 40% = 40 students
40% drink beer = 16
40% mixed = 16
20% both = 8

Group B 60% = 60
30% beer= 18
30% mixed = 18
20% both= 12

now we need both ( beer + mixed = both)

probability = total beer drinker = 16+18 =34 and both = 20
thus 20/34 = 10/17
Hence C

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