If Garry takes 3 pieces at random from the bag with the white chess pieces, what is the probability that he draws 2 pawns and a knight (in any order)? I stumbled upon this problem:
Garry keeps his white and black chess pieces in separate bags. For each color, there  are 8 pawns, 2 rooks, 2 bishops, 2 knights, 1 queen, and 1 king. If Garry takes 3  pieces at random from the bag with the white pieces, what is the probability that he  draws 2 pawns and a knight (in any order)?
I got 1/10.  Can anyone tell me if I'm right?
So, I just did pawn pawn knight, which is 8/16 x 7/15 x 2/14 = 1/30.  There are 3 unique combinations, so I multiplied by 3 and got 1/10.
Much appreciation <3
 A: Your solution is correct.
Here is another approach.  We wish to find the probability of selecting two of eight white pawns and one of two white knights while selecting three of the $8 + 2 + 2 + 2 + 1 + 1 = 16$ white chess pieces, which is
$$\frac{\dbinom{8}{2}\dbinom{2}{1}}{\dbinom{16}{3}} = \frac{28 \cdot 2}{560} = \frac{1}{10}$$
A: Yes.  Initially there are 16 pieces in the bag, 8 of which are pawns.  The probability the first piece drawn is a pawn is 8/16= 1/2.  There are then 15 pieces in the bag, 7 of which are pawns.  The probability the second piece drawn is a pawn is 7/15. There are then 14 pieces iplpn the bag, 2 of which are knights.  The probability the third piece drawn is a knight is 2/14= 1/7.  The probability the three pieces drawn are "pawn, pawn, knight", in that order, is (1/2)(7/15)(1/7)= 1/30.
It is easy to see that "two pawns and one knight" in any order, "pawn, pawn, knight", "pawn, knight, pawn", or "knight, pawn, pawn", are exactly the same: the   same fractions are multiplied in different orders.  So the probability of "pawn, pawn, knight", in any order, is 3(1/30)= 1/10.
That's what you did, right?
