Multiple choice

A box contains 10 different balls. Five balls are drawn simultaneously and then replaced and then seven balls are drawn. Then the probability that exactly three balls are common to the two draws is (p), then the value of (p) is:

  1. $\(\frac{1}{12}\)$
  2. $\(\frac{3}{12}\)$
  3. $\(\frac{5}{12}\)$
  4. $\(\frac{7}{12}\)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We need exactly 3 common balls between two draws. First draw selects 5 balls from 10. Second draw selects 7 balls from 10. Probability = [C(5,3) × C(5,4)] / C(10,7) = (10 × 5) / 120 = 50/120 = 5/12. The numerator counts ways to choose 3 common from first 5 and 4 new from remaining 5.

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