Question 1133521: A box contains 9 tickets numbered 1 to 9 inclusive. If three tickets are drawn from the box one at a time
without replacement, find the probability that they are alternately be odd, even, odd and even, odd, even
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! There are 9 tickets with 4 even numbered tickets and 5 odd numbered tickets
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a) odd, even, odd
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Probability(P) (odd, even, odd) = (5/9), (4/8), (4/7) = 80/504 = 10/63 is approximately 0.16
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b) even, odd, even
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P (even, odd, even) = (4/9) * (5/8) * (3/7) = 60/504 = 10/84 = 5/42 is approximately 0.12
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