Eight eligible bachelors and seven beautiful models happen randomly to have purchased single seats in the same 15-seat row of a theater. On the average, how many pairs of adjacent seats are ticketed for marriageable couples^{1}?

Let \(X\) be a random variable denoting the number of pairs that are marriageable. We can rewrite \(X\) as \(X = X_{1} + X_{2} + \ldots + X_{14}\), where each \(X_{i}\) is a random variable representing the number of marriageable couples in the \(i^{th}\) pair of adjacent seats. Note that each \(X_{i}\) can take on either \(0\) or \(1\) only.

We can easily calculate \(E[X_{i}]\) by considering the two arrangements that result in a marriageable pair, which are *Male-Female* and *Female-Male*:

\[E[X_{i}] = \left(\frac{8}{15}\right) \left(\frac{7}{14}\right) + \left(\frac{7}{15}\right) \left(\frac{8}{14}\right) = \frac{8}{15}\]

Then, applying the *Linearity of Expectation* property, we can calculate \(E[X]\) as follows:

\[\begin{split} E[X] &= E[X_{1} + X_{2} + \ldots + X_{14}] \\ &= E[X_{1}] + E[X_{2}] + \ldots + E[X_{14}] \\ &= 14\left(\frac{8}{15}\right) \end{split}\]

Thus, the expected number of marriageable pairs in the row of \(15\) seats is \(E[X] \approx 7.46\).

According to outdated cultural norms from the time of this problem’s publication.↩︎