Christine,

Preliminary idea
There are three mutually independent possibilities:

1. EXACTLY one got an A
2. EXACTLY two got an A
3. EXACTLY none got an A

Probability that AT LEAST ONE got an A
    = probability that EXACTLY one got an A + probability that EXACTLY two got an A
    = 1 - probability that none got an A

So, let's work on the probability that none got an A

Method 1 - selecting one person at a time
On the first selection, there are 12 out of 15 ways to select a person who did not get an A.

Because there is one less person, and this person didn't get an A: On the second selection, there are 11 out of 14 ways to select a person who did not get an A.

Thus, the probability that none got an A = (12/15) x (11/14) = (12x11)/(15x14) = 132/210 = 22/35

Method 2 - selecting two persons at the same time
The number of ways to select any 2 people out of 15
    = ₁₅C₂  (using theory of combinations - you may not be aware of this theory)

    = 15 choose 2
    = 15!/(13!x2!)
    = 15x14/2
    = 105

The number of ways to select 2 people from the 12 people who did not get an A
    = ₁₂C₂

    = 12!/(10!x2!)
    = 12x11/2
    = 66

The probability that none got an A = 66/105 = 22/35

By either method 1 or 2, the probability that none got an A = 22/35.

Hence, probability that AT LEAST ONE got an A = 1 - 22/35 = 13/35 (approximately 37%).

Paul Betts.