91Ó°ÊÓ

: Let's use the indicator variable approach. Let \(X_i\) denote the event where the \(i\)-th man has a woman on at least one side of him. Then, \(X_i\) can be either 0 or 1.

: We want to calculate the probability of \(X_i\) being 1, which is the probability that the \(i\)-th man has a woman on at least one side. \(P(X_i = 1) = 1 - P(X_i = 0)\) Let's find the probability that man \(i\) doesn't have a woman on either side, i.e., \(P(X_i = 0)\). There are 3 cases that a man can have other men on both sides: 1. The \(i\)-th man is at one of the ends of the line and has another man beside him. 2. The \(i\)-th man is in the middle and has men on both sides. 3. The \(i\)-th man is at one of the ends of the line and has no one beside him (this case is not practically possible, but we consider it for the calculation) Total number of possible cases = \(n+m\) Now let's find the probabilities for each case. 1. Probability of case 1: \(P(\text{case 1}) = \frac{2(n-1)}{n+m}\) 2. Probability of case 2: \(P(\text{case 2}) = \frac{(n-1)(n-2)}{(n+m)(n+m-1)}\) 3. Probability of case 3: \(P(\text{case 3}) = \frac{2}{n+m}\) Now add all the 3 cases probabilities and subtract it from 1 to find the probability of a man having a woman on at least one side. \(P(X_i = 1) = 1 - [P(\text{case 1}) + P(\text{case 2}) + P(\text{case 3})]= 1 - \frac{2(n-1) + (n-1)(n-2) + 2}{(n+m)(n+m-1)}\)

: Now we need to find the expected value, which is the expected number of men that have a woman on at least one side. We can use the formula for expectation: \(E(X) = \sum_{i=1}^n P(X_i = 1)\) Using the probability we found in step 2, and summing it up for \(n\) men, we have: \(E(X) = n \cdot [1 - \frac{2(n-1) + (n-1)(n-2) + 2}{(n+m)(n+m-1)}]\)

: The expected number of men that have a woman on at least one side is: \(E(X) = n \left[1 - \frac{2(n-1) + (n-1)(n-2) + 2}{(n+m)(n+m-1)}\right]\)