BTGmoderatorDC wrote: ↑

Sun Mar 28, 2021 5:39 pm

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


OA E

Solution:

There are 64 squares on a chessboard, of which 32 are red and 32 are black. Thus, the probability of randomly choosing one black square is 1/2, and the probability of randomly choosing one red square is also 1/2. If we are to choose 3 squares, the only outcomes that would NOT result in getting 1 red and 2 black or 2 red and 1 black would be getting all 3 red or all 3 black. We assume that we are choosing squares sequentially, so we use “sampling without replacement” in our approach.

P(3 red) = 32/64 x 31/63 x 30/62 = 5/42

P(3 black) = 32/64 x 31/63 x 30/62 = 5/42


P[(2 red/1 black) or (1 red/2 black)] = 1 - (5/42 + 5/42) = 1 - 10/42 = 32/42 = 16/21

Answer: E