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WAEC 2012 Mathematics · Question 34

A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are…

Question

A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours

  1. A.815\frac{8}{15}
  2. B.1325\frac{13}{25}
  3. C.1115\frac{11}{15}
  4. D.1315\frac{13}{15}

Answer: A. 815\frac{8}{15}

Explanation written with AI and checked against the answer key

Analyze the question: There are 10 balls (4 red, 6 black). Two are drawn without replacement. We want the probability that the two balls have different colours, meaning red then black, or black then red. Concept: For draws without replacement, the second probability changes because the total number of balls drops by 1. Add the probabilities of the two separate cases (mutually exclusive). Solution: Case 1: Red first, then black P = 4/10 × 6/9 = 24/90 Case 2: Black first, then red P = 6/10 × 4/9 = 24/90 Total: P(different colours) = 24/90 + 24/90 = 48/90 = 8/15 Check with combinations: Favourable = 4C1 × 6C1 = 24 Total = 10C2 = 45 P = 24/45 = 8/15 ✓ Answer: A. 8/15 Tip: For "different colours" with two draws, you can use the shortcut 2 × (a × b) / (n(n−1)) = 2×4×6/(10×9) = 48/90. Alternatively, use the combination method: (ways to pick one of each) ÷ (ways to pick any two). You can also check by computing 1 − P(same colour): P(both red) = 12/90, P(both black) = 30/90, so same = 42/90 and different = 48/90 ✓.

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