SOLUTION: 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 od
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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
:
a) odd, even, odd
:
Probability(P) (odd, even, odd) = (5/9), (4/8), (4/7) = 80/504 = 10/63 is approximately 0.16
:
b) even, odd, even
:
P (even, odd, even) = (4/9) * (5/8) * (3/7) = 60/504 = 10/84 = 5/42 is approximately 0.12
:
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