Question

A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red is:

• A. \( \frac{2}{91} \)
• B. \( \frac{3}{22} \)
• C. \( \frac{5}{71} \)
• D. \( \frac{7}{51} \)
• E.

Hint:

When calculating probability, use combinations to find the total possible outcomes and favorable outcomes.

Answer 1

The Correct Option is A

Step 1: Calculate the total number of balls \[ 4 \text{ (white)} + 5 \text{ (red)} + 6 \text{ (blue)} = 15 \text{ balls}. \] Step 2: Compute the total ways to choose 3 balls from 15 \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455. \] Step 3: Compute the ways to choose 3 red balls from 5 \[ \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10. \] Step 4: Compute the probability of drawing 3 red balls \[ \frac{\binom{5}{3}}{\binom{15}{3}} = \frac{10}{455} = \frac{2}{91}. \] Thus, the probability of drawing 3 red balls is \( \frac{2}{91} \).