If you have 15 women and 15 men at a party, given that there are 15 available tables, what is the expected number of tables which will contain both a man and a woman. Note: Each table can contain an unlimited amount of people.

  • $\begingroup$ Does each person have to sit down? $\endgroup$ – Jonathan Davidson Jul 17 '17 at 5:18
  • $\begingroup$ @JonathanDavidson yes they do. $\endgroup$ – Math2Hard Jul 17 '17 at 5:34
  • $\begingroup$ You need to make some assumption about how seats are chosen. Otherwise all sorts of scenarios are possible. $\endgroup$ – Robert Israel Jul 17 '17 at 6:05

Sketch of proof: Let $X_i$ be the random variable which is $1$ if there is at least one woman and one man at table $i$, and $0$ otherwise. The variable you want the expectation of is $$ X_1+X_2+\cdots+X_{15} $$ Now remember that expectation is linear.

  • $\begingroup$ What about the men? $\endgroup$ – Robert Israel Jul 17 '17 at 6:08
  • $\begingroup$ @RobertIsrael I missed that in the question. Easily fixed. Still, my main point, that you can focus on a single table and not worry about interactions, stands. $\endgroup$ – Arthur Jul 17 '17 at 6:09
  • $\begingroup$ And why should $E(X_j)$ all be the same? $\endgroup$ – Robert Israel Jul 17 '17 at 6:13
  • $\begingroup$ Maybe Table $1$ is the only one with a good view of the football game, so all the men will sit there. $\endgroup$ – Robert Israel Jul 17 '17 at 6:17
  • $\begingroup$ @RobertIsrael Doesn't matter. Expectation is linear even with dependent, differently-distributed variables. You just have to calculate $E(X_1)$ on its own, so the answer is not $15E(X_1)$, but $E(X_1)+14E(X_2)$. What I said in my answer still stands, and is, in my opinion, the easiest route to the answer by a wide margin even if all the tables were different, and even if some people wanted to sit together or apart. $\endgroup$ – Arthur Jul 17 '17 at 6:31

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