tried to solve this a different way but encounter a different result. Please look
P(p(1st freshman AND 2nd sophomore) OR p(1st sophomore AND 2nd freshman)
=p1 + p2
=(2/8 * 2/7) + (2/8 * 2/7) (7 because one person already choosen, so OUT)
=1/14 + 1/14
This is different from OA. Anyone can point out the problem with solution above?
I'll give it a try.
I think what you want is P(F1 & So1) + P(F1 & So2) + P(F2 + So1) + P(F2 + So2). There are four possible desireable outcomes. Since there are 24 possible outcomes, I get 4/24, which is 1/6.
Doing it your way, I think you would want to do it like this:
[P(choosing a freshman first) + P(choosing a sophomore second)] + [P(choosing a sophomore first) + P(choosing a freshman second)] =
(1/4 x 2/6) + (1/4 + 2/6) = 4/24 = 1/6
Your denominator was 7, when it should have been 6, because there was a requirement that the two students be from different grades, so once a student is chosen, the other student in his/her grade can't be chosen.