Difficult Math Problem #58 - Probability

[800guy]

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

[aim-wsc]

for first years
ratio of
beer to mixed drinks to both drinks
is
40:40:20
=4:4:2

for 2nd years
it is 30:30:20...................(rest of the 20% do not drink here's a catch)
=3:3:2

also we know that 60% of pub members are 2nd years
therefore
1st years===(4:4:2) 4==>16:16:8

2nd years===(3:3:2) 6 ==>18:18:12

so in total ===========>34:34:20
is the ratio of
beer to mixed drinks to both drinks


so if you pick any random person who drinks beer (among 34x) the probability that the person also drinks mixed drinks ( among 20x)
will
be

20/34 ie
Answer:10/17

[800guy]

OA:

The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

The total number of possible outcomes is the total percent of students drinking beer.

40% of the students are first year students. 40% of those students are drinking beer. Thus, the first years drinking beer make up (40% * 40%) or 16% of the total number of students.

60% of the students are second year students. 30% of those students are drinking beer. Thus, the second years drinking beer make up (60% * 30%) or 18% of the total number of students.

(16% + 18%) or 34% of the group is drinking beer.

The outcomes that result in A is the total percent of students drinking beer and mixed drinks.

40% of the students are first year students. 20% of those students are drinking both beer and mixed drinks. Thus, the first years drinking both beer and mixed drinks make up (40% * 20%) or 8% of the total number of students.

60% of the students are second year students. 20% of those students are drinking both beer and mixed drinks. Thus, the second years drinking both beer and mixed drinks make up (60% * 20%) or 12% of the total number of students.

(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.

[AbhiJ]

I think the answer should be 20/(20+34), as the people drinking beer would include people drinking both drinks.