91Ó°ÊÓ

Let's define an indicator random variable \(X_i\) for each wife, where \(i = 1, 2, ..., 10\), such that: \(X_i = \begin{cases}1 & \text{if wife } i \text{ is seated next to her husband} \\ 0 & \text{otherwise}\end{cases}\)

We can calculate the expected number of wives seated next to their husbands as the sum of the expected value of each indicator random variable \(X_i\). The expected value of the sum of the indicator random variables can be written as: \(E[\sum_{i=1}^{10} X_i] = \sum_{i=1}^{10} E[X_i]\) Now we need to determine \(E[X_i]\) for a single wife. Since each wife has a probability of \(\frac{1}{19}\) of being seated next to her husband (because there are 19 possible choices for her to sit, and only 1 of them is next to her husband), we have: \(E[X_i] = 1 \cdot \frac{1}{19} + 0 \cdot \frac{18}{19} = \frac{1}{19}\) Now, we can calculate the expected value: \(E[\sum_{i=1}^{10} X_i] = 10 \cdot \frac{1}{19} = \frac{10}{19}\)

Now let's calculate the variance of the number of wives seated next to their husbands. The variance for the sum of the indicator random variables can be written as: \(Var(\sum_{i=1}^{10} X_i) = \sum_{i=1}^{10} Var(X_i) + \sum_{i \neq j} Cov(X_i, X_j)\) Since the random variables \(X_i\) are not independent (if wife 1 sits next to her husband, it affects the probability for the other wives), we need to find the covariance between \(X_i\) and \(X_j\) where \(i \neq j\): \(Cov(X_i, X_j) = E[X_i X_j] - E[X_i]E[X_j]\) Here, \(E[X_i X_j] = P(X_i = 1 \text{ and } X_j = 1) = \frac{1}{18}\) because, if wife \(i\) sits immediately next to her husband, then there are 18 possible places for wife \(j\) to sit such that she also sits next to her husband. So, we have \(Cov(X_i, X_j) = \frac{1}{18} - \frac{1}{19}\frac{1}{19} = \frac{1}{18} - \frac{1}{361}\). Now back to calculating the variance, we have: \(Var(\sum_{i=1}^{10} X_i) = \sum_{i=1}^{10} Var(X_i) + \sum_{i \neq j} Cov(X_i, X_j)\) \(Var(\sum_{i=1}^{10} X_i) = 10 \cdot Var(X_i) + \binom{10}{2} \cdot Cov(X_i, X_j)\) The variance \(Var(X_i)\) can be calculated by \(Var(X_i) = E[X_i^2] - E[X_i]^2 = \frac{1}{19} - \frac{1}{361}\), so \(Var(\sum_{i=1}^{10} X_i) = 10 \cdot (\frac{1}{19} - \frac{1}{361}) + \binom{10}{2} \cdot (\frac{1}{18} - \frac{1}{361})\) \(Var(\sum_{i=1}^{10} X_i) \approx 2.127\) So, the expected number (a) of wives seated next to their husbands is \(\frac{10}{19}\) and the variance (b) is approximately 2.127.