m00li's post in two pawns was marked as the answer
April 21, 2014

there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black.
Method 1:
A = number of ways of selecting 5 out of the above
B = number of ways of selecting 5 such that no pawn is selected
C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected
then Answer = (A-C)/(A-B)
A = coeff of x5 in (1-x9)2(1-x3)6(1+x)4(1-x)-8 = 2540
B = coeff of x5 in (1-x3)6(1+x)4(1-x)-6 = 876
C = coeff of x5 in (1+x)(1-x9)(1-x3)6(1+x)4(1-x)-7=2230

Answer = 310/1664 = 0.186298

Method 2
A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected
B = no. of ways of selecting 5
C = no. of ways of selecting 5 without any pawns
Answer = A/(B-C)