Multiple choice

A bag contains 4 white, 5 red and 6 black balls. Three are drawn at random. Find the probability that (i) no ball drawn is black, (ii) exactly 2 are black (iii) all are of the same colour.

  1. $\displaystyle \frac{12}{65},\frac{27}{91},\frac{6}{91}$
  2. $\displaystyle \frac{12}{65},\frac{27}{91},\frac{24}{455}$
  3. $\displaystyle \frac{12}{65},\frac{27}{91},\frac{14}{455}$
  4. $\displaystyle \frac{12}{65},\frac{27}{91},\frac{34}{455}$
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D Correct answer
Explanation

Total balls = 15. Ways to choose 3 = 15C3 = 455. (i) No black: Choose 3 from 9 (4 white + 5 red) = 9C3 = 84. Prob = 84/455 = 12/65. (ii) Exactly 2 black: Choose 2 from 6 black and 1 from 9 others = 6C2 * 9C1 = 15 * 9 = 135. Prob = 135/455 = 27/91. (iii) Same color: 4C3 + 5C3 + 6C3 = 4 + 10 + 20 = 34. Prob = 34/455.

AI explanation

The total number of balls is 15, so the total number of ways to draw 3 balls is 15C3 = 455. The probability of drawing no black balls is choosing 3 from the 9 non-black balls, giving 9C3/455 = 84/455 = 12/65. The probability of drawing exactly 2 black balls and 1 non-black ball is (6C2 times 9C1)/455 = (15 times 9)/455 = 135/455 = 27/91. The probability that all 3 are the same colour is the sum of drawing 3 white, 3 red, or 3 black balls, calculated as (4C3 + 5C3 + 6C3)/455 = (4 + 10 + 20)/455 = 34/455.

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