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A container holds 13 red, 14 green, and 15 white balls. Let p1 denote the probability that exactly 2 white balls are obtained when 4 balls are selected. Now the number of balls of each colour is doubled. Let p2 denote the probability that exactly 4 white balls are obtained when 8 balls are selected. Then
- p1 = p2
- p1 > p2
- p1 < p2
- None of these
Correct answer: p1 > p2
Solution
p1 = C(15,2)*C(27,2)/C(42,4) ≈ 0.329. After doubling: p2 = C(30,4)*C(54,4)/C(84,8) ≈ 0.199. Since 0.329 > 0.199, we have p1 > p2.
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