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Question

A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box without replacement. The probability that they are alternatively either {odd, even, odd} of {even, odd, even} is

A
516
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B
517
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C
518
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D
417
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Solution

The correct option is C 518
First find probability that the 3 tickets drawn are ( odd, even, odd )

There are 9 tickets numbered 1 to 9
So, initially there are 5 odd numbered tickets and 4 even numbered.

The probability that first ticket drawn is odd numbered =59

Now, the second ticket is drawn without replacement. There are 8 tickets in the box, out of which 4 are odd numbered and 4 are even.

The probability that second ticket drawn is even numbered =48

Now, the third ticket is drawn without replacement

Probability that third ticket drawn is odd numbered =47

probability that the 3 tickets drawn are ( odd, even, odd )
=59×48×47=1063


Second case,
the 3 tickets drawn are ( even, odd, even )

There are 9 tickets numbered 1 to 9
So, initially, there are 5 odd numbered tickets and 4 even numbered.

The probability that the first ticket drawn is even numbered =49

Now, the second ticket is drawn without replacement. There are 8 tickets in the box, out of which 5 are odd numbered and 3 are even.

The probability that the second ticket was drawn is odd numbered =58

Now, the third ticket is drawn without replacement

The probability that the third ticket drawn is even numbered =37

The probability that the 3 tickets drawn are ( even, odd, even )
=49×58×37=542

Probability that the 3 tickets drawn are either ( even, odd, even ) or
( odd, even, odd ) =1063+542=518

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