4. A box contains 10 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.
a. Odd, even, odd
b. Even, odd, even
correct answer plss
a. Odd, even, odd
b. Even, odd, even
correct answer plss
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Question:
There are 9 tickets with 4 even numbered tickets and 5 odd numbered tickets
Answer:
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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First find probability that the 3 tickets drawn are ( odd, even, odd )
So, initially there are 5 odd numbered tickets and 4 even numbered.
9
5
8
4
=
9
4
×
8
5
×
7
3
=
42
5
( odd, even, odd ) =
63
10
+
42
5
=
18
5