ykaiim 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
I didn't understand what the question asked
. I tried to calculate but all mismatch. Please provide some explanations.
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