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, 50% are drinking beer, 20% are drinking mixed drinks, and 30% 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? Options
A) 1 B) 4/7 C) 10/19 D) 7/24 E) 7/10
Suppose there are 100 students   Group A : 40% = 40 students   50% drink beer = 20   30% mixed = 12   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 = 20+18= 
<gadget id="calculator">20+18</gadget>
<output>38</output>
38 and both= 20   thus 20/38 = 10/19   Hence 10/19
<result>C</result>
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, 50% are drinking beer, 20% are drinking mixed drinks, and 30% 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?