A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

Solution

There are 9 tickets numbered 1 through 9, with 5 odd and 4 even numbers. Three tickets are drawn without replacement, and the order matters.

The total number of possible outcomes is the permutation $P(9,3) = 9 \times 8 \times 7 = 504$.

The favorable outcomes are sequences alternating odd and even: either odd–even–odd or even–odd–even.

For the odd–even–odd pattern, the number of outcomes is $5 \times 4 \times 4 = 80$:
Choose an odd ticket from 5, then an even from 4, then another odd from the remaining 4 odd tickets.

For the even–odd–even pattern, the number of outcomes is $4 \times 5 \times 3 = 60$:
Choose an even ticket from 4, then an odd from 5, then another even from the remaining 3 even tickets.

The total favorable outcomes are $80 + 60 = 140$.
The probability is $\frac{140}{504} = \frac{5}{18}$.

$\boxed{\frac{5}{18}}$