aragonn wrote:
Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:
A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21
When looking for probability, don't get worked up on probability and combinations..
Just be sure that you choose same for both favorable and total.(I) combinations, that is order does not matter..Favorable outcomes - 32C2 for first color and 32C1 for second, and both colors can be used twice, so 32C2*32C1*2=\(\frac{32*31*32*2}{2}\)
Total outcomes - 64C3 =\(\frac{64*63*62}{3*2}\)
Thus, Probability = \(\frac{32C2*32C1*2}{64C2}=\frac{32*31*32*3}{64*63*62}=\frac{16}{21}\)
(II) Permutations, that is order matters..Favorable outcomes - 32P2 for first color and 32P1 for second, and both colors can be chosen in \(\frac{3!}{2}=3\) ways, so 32*31*32*3
Total outcomes - 64P3 =\(64*63*62\)
Thus, Probability =\frac{32*31*32*3}{64*63*62}=\frac{16}{21}[/m]
E
Permutations - Why 2 colors can be chosen in 3!/2? Please explain. Thank you.