A team has 10 players named A, B, C, ... . Before each game an o®ensive

captain, a defensive captain, and a water-boy are determined by chance. No one may have
more than one of these honors per game.

The probability the C and D both win honors for the sixth game is closest to?

The probability that for the third game E wins honors but B does not is
closest to?

During a season with 16 games, the number of times that A and F together
can expect to win honors as water-boy is closest to?

Are you "answer mooching" or don't you understand anything about probabilities?

Please read the chapter on binomial distributions, then try the problems and post your work.

To calculate the probabilities, we need to determine the number of possible outcomes for each game.

1. The probability that C and D both win honors for the sixth game:

This can occur if either C or D becomes the offensive captain and the other becomes the defensive captain. Since no one can have more than one honor per game, the water-boy will be one of the remaining 8 players.

The total number of possibilities for the sixth game is the product of the choices for offensive captain, defensive captain, and water-boy:
10 choices for offensive captain x 9 choices for defensive captain x 8 choices for water-boy = 720 possibilities.

Out of these, the favorable outcomes are when C becomes either the offensive or defensive captain, and D takes the remaining honor. This can happen in two ways: C becomes the offensive captain and D becomes the defensive captain or vice versa.

Therefore, the probability that C and D both win honors for the sixth game is 2/720, which simplifies to 1/360.

2. The probability that for the third game E wins honors, but B does not:

In this case, E can have any of the three honors, and B cannot have any honor. There are 10 choices for the offensive captain, 9 choices for the defensive captain, and 8 choices for the water-boy.

Therefore, the total number of possibilities for the third game is 10 x 9 x 8 = 720.

Out of these, the favorable outcomes occur when E gets an honor (any of the three) and B is not chosen for any honor.

For E to get an honor, we have 3 choices (offensive captain, defensive captain, or water-boy) for E.

For B not to get any honor, we have 7 choices (excluding B from the 3 honors and the remaining 6 players to choose from).

Hence, the probability that for the third game E wins honors, but B does not, is (3 x 7)/720, which simplifies to 7/240.

3. The number of times A and F together can expect to win honors as water-boys in a season with 16 games:

Since there are 16 games, we will multiply the probability of A and F winning the water-boy honor in a single game by the total number of games.

In a single game, there are 10 choices for the offensive captain, 9 choices for the defensive captain, and 8 choices for the water-boy.

For A and F to both win the water-boy honor, we need A to be a water-boy and F to be one of the remaining 7 players.

Thus, we have 10 choices for A, 7 choices for F, and 8 choices for the remaining player in each of the 16 games.

So the total number of possibilities for A and F winning the water-boy honor in a season is (10 x 7 x 8)^(16 games), which can be calculated using the appropriate exponential function.

This will give the number of times A and F can expect to win honors as water-boys in a season with 16 games.

Note: The closest value is an estimation, as we are dealing with probabilities and the exact values may involve long calculations.