I am having trouble understanding how to solve this problem. I found an answer, about 74.2%, but I am not sure if that is correct. Any help would be greatly appreciated!

A chess board has 64 squares,
32 white and 32 black, and is played
with 16 black and 16 white pieces. If all the pieces are placed randomly on
the board, what is the probability of two
white knights being on black squares or
a black bishop being on a black square?

To solve this problem, we need to calculate the probability of two white knights being on black squares and the probability of a black bishop being on a black square, and then add these probabilities together.

Let's first calculate the probability of two white knights being on black squares.

The probability of the first white knight being on a black square is 32/64 since there are 32 black squares out of 64 total squares.

After placing the first white knight on a black square, the probability of the second white knight also being on a black square is slightly different. We need to take into account that there are now 63 squares left, and only 31 of them are black since the first white knight took one black square. Therefore, the probability of the second white knight being on a black square is 31/63.

To calculate the probability of both white knights being on black squares, we multiply the individual probabilities:

P(white knights on black squares) = (32/64) * (31/63) = 992/4032 ≈ 0.2457 ≈ 24.57%

Next, let's calculate the probability of a black bishop being on a black square.

The probability of a black bishop being on a black square is 32/64 since there are 32 black squares out of 64 total squares.

Now we can add the two probabilities to get the final result:

P(two white knights on black squares or black bishop on black square) = P(white knights on black squares) + P(black bishop on black square)
= 992/4032 + 32/64
= 992/4032 + 32/64
= 992/4032 + 2/64
= 992/4032 + 64/4032
= 1056/4032 ≈ 0.2619 ≈ 26.19%

Therefore, the probability of two white knights being on black squares or a black bishop being on a black square is approximately 26.19%, which is not the same as 74.2%. So the answer you found, about 74.2%, is incorrect.

To find the probability of the given condition, we need to determine the total number of favorable outcomes and the total number of possible outcomes.

Let's first calculate the total number of favorable outcomes:

1. Two white knights on black squares:
- The total number of black squares on a chessboard is 32.
- To place two white knights on black squares, we have to choose 2 out of the 16 white knights and place them on 2 out of the 32 black squares.
- The number of ways to choose 2 out of 16 is given by the combination formula: C(16, 2) = 16! / (2!(16-2)!) = 16! / (2!14!) = (16 * 15) / (2 * 1) = 120.
- The number of ways to choose 2 out of 32 is given by the combination formula: C(32, 2) = 32! / (2!(32-2)!) = 32! / (2!30!) = (32 * 31) / (2 * 1) = 496.
- So, the total number of favorable outcomes for two white knights on black squares is 120 * 496 = 59,520.

2. A black bishop on a black square:
- The total number of black squares on a chessboard is 32.
- To place one black bishop on a black square, we have to choose 1 out of the 16 black bishops and place it on 1 out of the 32 black squares.
- The number of ways to choose 1 out of 16 is given by the combination formula: C(16, 1) = 16! / (1!(16-1)!) = 16! / (1!15!) = 16.
- So, the total number of favorable outcomes for a black bishop on a black square is 16 * 32 = 512.

Now, let's calculate the total number of possible outcomes:

- The total number of ways to place 16 white pieces on 64 squares is given by the permutation formula: P(64, 16) = 64! / (64-16)! = 64! / 48! = 7.95 x 10^19.
- So, the total number of possible outcomes is 7.95 x 10^19.

Finally, let's calculate the probability:

- Probability = (Number of favorable outcomes) / (Number of possible outcomes)
- Probability = (59,520 + 512) / (7.95 x 10^19)
- Probability ≈ 7.47 x 10^-18

Therefore, the probability of two white knights being on black squares or a black bishop being on a black square is approximately 7.47 x 10^-18.