Dse question (Amit sir plz solve)

[bhavya jain]

was anyone able to solve this question?
1) 5 men and 5 women are seated randomly in a single row of chairs. The expected number of women sitting next to atleast 1 man equals
a) 11/3
b) 13/3
c) 35/9
d) 37/9

2) 5 men and 5 women are seated randomly in a single circle of chairs. The expected number of women sitting next to atleast 1 man equals
a) 23/6
b) 25/6
c) 4
d) 17/4

[Noel]

Call the women 1, 2, 3, 4, and 5.

Let Xi=1 if the woman named i is seated next to at least one man, and let Xi=0 otherwise.

We want E(X1+⋯+X5), which by linearity of expectation is E(X1)+⋯+E(X5).

To find E(Xi), calculate Pr(Xi=1). It is easier to first find the probability that Xi=0. Suppose woman i is the first to be seated. What is the probability that both of her neighbours are women?

Remark: Note that we can use exactly the same method if there are w women and m men.

The method of indicator random variables can be quite useful. It bypasses the sometimes very difficult problem of finding the probability distribution of the random variable whose expectation we are seeking.

[Amit Goyal]

Hi Bhavya,

Noel is absolutely right. Thats one of the best ways to do these problems. The detailed solutions are attached.

[Amit Goyal]

[L]

Amit Sir, isn't for circular permutations, total permutations are (n-1)!. So, in second question, while calculating probability, denominator should be 9! and numerator should be same. But, that would make the expression -ve. I am not able to understand where I am getting it wrong. Please help!

[Ansuman7]

In circular arrangement if seats are numbered i think it is n! not (n-1)!.