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A bag contains coupons numbered 1 through 15. Three coupons are drawn at random. Given that the sum of the three numbers drawn is odd, what is the probability p that all three numbers are odd? Find the value of 4p.
- 1
- 2
- 3
- 4
Correct answer: 2
Solution
P(sum odd) uses cases: all 3 odd or exactly 1 odd. P(all odd | sum odd) = C(8,3) / [C(8,3) + C(8,1)*C(7,2)] = 56 / [56 + 168] = 56/224 = 1/4. So 4p = 1... let me recount.
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