Question:
A certain junior class class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

(A) 3/40,000
(B) 1/3,600
(C) 9/2,000
(D) 1/60
(E) 1/15

Answer:
(A)

I looked at this problem as having 2 winning scenarios:

(1) 1 Junior, 1 Senior
(2) 1 Senior, 1 Junior

In (1), the "winning scenario" the first "winning event" looks at the probability that the student from the junior class (1,000 students) is a member of a sibling pair (60 pairs). This results in a P = 60/1,000. Then, the second "winning event" looks at the probability that the student from the senior class (800 students) is the other member of the pair. This results in a P = 1/800. The probability of this "winning scenario" is (60/1,000)*(1/800) = 3/40,000.

My question is how come we don't also look at the 2nd winning scenario (2), since a senior could be picked first, then a junior? Then, the probability of either event chain - (1) or (2) occurring would be the sum of these individual probabilities?

If you calculate the second scenario, you'll get the same answer: 3/40000.

Each scenario already accounts for choosing both of the siblings - it doesn't matter which you happen to choose first. You've calculated the number for choosing 1 student AND that student's sibling, period. The order in which you choose only matters if the problem specifies such (and then the calculations can become more complicated).