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

Step-by-step solution

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 8 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 4 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–ev