A bag contains 5 white, 7 black, and 4 red balls. Three balls are drawn one after another without replacement. What is the probability that the sequence of balls drawn is white, black, and red?

Instruction: A bag contains 5 white, 7 black and 4 red balls. Draw three balls uniformly without replacement. Find the probability of white first, black second and red third, in that exact order.

Context: This question tests the candidate's understanding of conditional probability and the complexities introduced by drawing without replacement.

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Official answer

There are 16 balls initially. After drawing a white ball, 15 remain, including all 7 black balls. After drawing a black ball, 14 remain, including all 4 red balls.

P(white, then black, then red) = (5/16)(7/15)(4/14) = 1/24 ≈ 4.1667%.

These are conditional probabilities, so independence is not required. The denominators shrink because the draws are without replacement.

If the question instead asked for one of each color in any order, there would be six disjoint orders with the same probability, giving 6/24 = 1/4. Do not apply that factor when the order is specified.

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