Questions: A committee of 3 members is selected from a club made up of 11 junior members and 19 senior members. What is the expected number of juniors in the committee? Fill out the probability table below, where x represents the number of juniors in the committee. x 0 1 2 3 P(x) (Type integers or simplified fractions.)

Solution Steps

To find the probabilities of having \( k \) juniors in a committee of 3 members selected from a club with 11 junior members and 19 senior members, we use the hypergeometric distribution formula:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

where:

• \( N = 30 \) (total members),
• \( K = 11 \) (junior members),
• \( n = 3 \) (committee size),
• \( k \) is the number of juniors in the committee.

Calculating for each possible value of \( k \):

1. For \( k = 0 \): \[ P(X = 0) = \frac{\binom{11}{0} \binom{19}{3}}{\binom{30}{3}} = 0.2387 \]

2. For \( k = 1 \): \[ P(X = 1) = \frac{\binom{11}{1} \binom{19}{2}}{\binom{30}{3}} = 0.4633 \]

3. For \( k = 2 \): \[ P(X = 2) = \frac{\binom{11}{2} \binom{19}{1}}{\binom{30}{3}} = 0.2574 \]

4. For \( k = 3 \): \[ P(X = 3) = \frac{\binom{11}{3} \binom{19}{0}}{\binom{30}{3}} = 0.0406 \]

The expected number of juniors in the committee is calculated using the formula for the mean of a discrete probability distribution:

\[ \text{Mean} = E(X) = \sum_{k=0}^{3} k \cdot P(X = k) \]

Substituting the probabilities calculated:

\[ E(X) = 0 \times 0.2387 + 1 \times 0.4633 + 2 \times 0.2574 + 3 \times 0.0406 = 1.1 \]

The variance \( \sigma^2 \) is calculated as follows:

\[ \text{Variance} = \sigma^2 = \sum_{k=0}^{3} (k - E(X))^2 \cdot P(X = k) \]

Calculating each term:

\[ \sigma^2 = (0 - 1.1)^2 \times 0.2387 + (1 - 1.1)^2 \times 0.4633 + (2 - 1.1)^2 \times 0.2574 + (3 - 1.1)^2 \times 0.0406 = 0.649 \]

The standard deviation \( \sigma \) is then:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{0.649} \approx 0.805 \]

Final Answer