Multiple choice

A box contains 2 red, 3 blue and 4 black balls. $3$ balls are drawn at random from the box. $p$ is the probability for the balls to be of different colours; $q$ is the probability for $2$ balls of the same colour and the third of different colour; $r$ is the probability for all the $3$ balls to be of the same colour. Arrange them in the decreasing order of magnitude

  1. $p, q, r$
  2. $q, r, p$
  3. $q, p, r$
  4. $r, q, p$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total balls = 9. Total ways to draw 3 = 9C3 = 84. p (diff colors) = (2C1 * 3C1 * 4C1)/84 = 24/84 = 2/7 approx 0.28. r (same color) = (2C3 + 3C3 + 4C3)/84 = (0 + 1 + 4)/84 = 5/84 approx 0.06. q (2 same, 1 diff) = 1 - p - r = 1 - 24/84 - 5/84 = 55/84 approx 0.65. Thus q > p > r.

AI explanation

The total number of ways to draw 3 balls from 9 is 84. For different colours, we calculate 2C1 * 3C1 * 4C1 = 24, giving p = 24/84. For two of the same colour and one different, we calculate the combinations for each color pair, totaling 46, which gives q = 46/84. All three being the same is impossible since no colour has 3 balls, so r = 0. Therefore, the decreasing order of magnitude is q, p, r.

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