Alchemist1320 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

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