Questions: A football team consists of 12 each freshmen and sophomores, 20 juniors, and 13 seniors. Four players are selected at random to serve as captains. Find the probability of the following. Enter your answers as fractions or as decimals rounded to 4 decimal places. Part 1 of 4 (a) All 4 are seniors. P(4 seniors)=0.0018 Part 2 of 4 (b) There is 1 each: freshman, sophomore, junior, and senior. P(1 of each)=0.0948 Part 3 of 4 (c) There are 2 sophomores and 2 freshmen.

Solution Steps

To solve these probability problems, we need to use combinations to determine the number of ways to select players and then calculate the probability based on these combinations.

1. Calculate the total number of ways to choose 4 players out of the total number of players.
2. Calculate the number of ways to choose 4 seniors out of the total number of seniors.
3. Divide the number of ways to choose 4 seniors by the total number of ways to choose 4 players.

1. Calculate the total number of ways to choose 4 players out of the total number of players.
2. Calculate the number of ways to choose 1 freshman, 1 sophomore, 1 junior, and 1 senior.
3. Divide the number of ways to choose 1 of each by the total number of ways to choose 4 players.

1. Calculate the total number of ways to choose 4 players out of the total number of players.
2. Calculate the number of ways to choose 2 sophomores out of the total number of sophomores and 2 freshmen out of the total number of freshmen.
3. Divide the number of ways to choose 2 sophomores and 2 freshmen by the total number of ways to choose 4 players.

The total number of players is: \[ 12 + 12 + 20 + 13 = 57 \] The total number of ways to choose 4 players out of 57 is: \[ \binom{57}{4} = 395010 \]

The number of ways to choose 4 seniors out of 13 is: \[ \binom{13}{4} = 715 \] The probability of all 4 being seniors is: \[ P(\text{4 seniors}) = \frac{715}{395010} \approx 0.0018 \]

The number of ways to choose 1 freshman, 1 sophomore, 1 junior, and 1 senior is: \[ \binom{12}{1} \times \binom{12}{1} \times \binom{20}{1} \times \binom{13}{1} = 12 \times 12 \times 20 \times 13 = 37440 \] The probability of 1 each (freshman, sophomore, junior, senior) is: \[ P(\text{1 each}) = \frac{37440}{395010} \approx 0.0948 \]

The number of ways to choose 2 sophomores out of 12 and 2 freshmen out of 12 is: \[ \binom{12}{2} \times \binom{12}{2} = 66 \times 66 = 4356 \] The probability of 2 sophomores and 2 freshmen is: \[ P(\text{2 sophomores, 2 freshmen}) = \frac{4356}{395010} \approx 0.0110 \]

Final Answer