Question

An urn contains 3 black and 5 red balls. If 3 balls are drawn at random from the urn, the mean of the probability distribution of the number of red balls drawn is:

• A. \( \frac{45}{28} \)
• B. \( \frac{15}{8} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{3}{7} \)

Hint:

When calculating expected values, multiply each outcome by its probability and sum them up. This helps you find the mean of the distribution.

Answer 1

The Correct Option is B

Step 1: Understand the problem. The problem is related to finding the mean of the probability distribution for the number of red balls drawn when 3 balls are drawn from an urn containing 3 black and 5 red balls. The number of red balls that can be drawn ranges from 0 to 3, and we need to calculate the expected value (mean) of this distribution. 
Step 2: Calculating probabilities. The number of possible outcomes when drawing 3 balls from the urn is: \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. \] Next, we calculate the probability of drawing 0, 1, 2, and 3 red balls. - For 0 red balls, all 3 balls must be black. The number of ways to choose 3 black balls is: \[ \binom{3}{3} = 1. \] So the probability is: \[ P(0 \text{ red balls}) = \frac{1}{56}. \] - For 1 red ball, we need to choose 1 red ball and 2 black balls. The number of ways to do this is: \[ \binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15. \] So the probability is: \[ P(1 \text{ red ball}) = \frac{15}{56}. \] - For 2 red balls, we need to choose 2 red balls and 1 black ball. The number of ways to do this is: \[ \binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30. \] So the probability is: \[ P(2 \text{ red balls}) = \frac{30}{56}. \] - For 3 red balls, all 3 balls must be red. The number of ways to choose 3 red balls is: \[ \binom{5}{3} = 10. \] So the probability is: \[ P(3 \text{ red balls}) = \frac{10}{56}. \] Step 3: Calculating the expected value. The expected value (mean) of the number of red balls drawn is the sum of each outcome multiplied by its probability: \[ E(X) = 0 \times \frac{1}{56} + 1 \times \frac{15}{56} + 2 \times \frac{30}{56} + 3 \times \frac{10}{56}. \] Simplifying: \[ E(X) = \frac{0 + 15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}. \] Thus, the mean of the probability distribution of the number of red balls drawn is \( \frac{15}{8} \).