This is a combination problem. There are 6 people from which 2 will be selected. This means that the total number of ways to pick people is a combination of 2 from 6: 6!/(2!4!) = 15.

Of these 15 possible group, only one group of 2 will include both Joshua and Jose. Therefore the probability that they are both picked is 1/15

The correct answer is A. A more detailed explanation as well as video solution can be viewed at GMATPrep Question 1208. To practice similar questions, set topic='combinatorics' and difficulty='600-700 & 700+' in the Drill Generator

Good luck,
-Patrick

gmatruler wrote:Joshua and jose work at an auto repair center with 4 other workers. For a survey on healthcare insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

A) 1/15
B) 1/12
C) 1/9
D) 1/6
E) 1/3

We also could handle this as a straight probability question.

Probability = (good outcomes)/(total outcomes)

In this problem:

good outcome = Josh or Jose
total outcomes = total number of people

Probability of picking Josh or Jose on the first pick: 2/6 (because there are 2 good outcomes out of a total of 6 people)

Probability of picking Josh or Jose on the second pick: 1/5 (because if Josh or Jose was chosen on the first pick, then only 1 good outcome is left for the 2nd pick, out of a total of 5 remaining people)

We need both of these events to happen. When we want multiple events to happen, we multiply the fractions:

P(A and B) = P(A) * P(B)

We multiply the fractions because the more events we want to happen together, the smaller the probability, and when we multiply fractions, the result just keeps getting smaller.

2/6 * 1/5 = 2/30 = 1/15.

The correct answer is A.

Wow 3 different, expert ways to solve. Thank you