There are $2m$ persons forming $m$ couples who live together at a given time. Suppose that at some later time, the probability of each person being alive is $p$, independently of other persons. At that later time, let $A$ be the number of persons that are alive and let $S$ be the number of couples in which both partners are alive. For any number of total surviving persons $a$, find $E[S∣A=a]$.

I don't have the answer to the problem. Here is my proposed solution, can anyone verify whether it is correct.

Let $$(S| A= a) = (S_1| A = a) + (S_2| A = a) + ... + (S_m | A = a)$$

Here $(S_i| A = a)$ is a indicator variable indicating whether both members in a couple are alive or not. So the probability of $p(S_i | A = a)$ is given by $$\frac{p(S_i \cap A = a)}{p(A = a)}$$ where the numerator is given by $p^2 {n - 2 \choose a-2}p^{a-2}(1-p)^{n-2-(a-2)}$ and denominator is given by ${n \choose a}p^a(1-p)^{n-a}$. Therefore $E[S_i | A = a]$ is now known to us, and by symmetry all the indicator variables will have same expectation. So we have the final answer as $m \cdot \frac{p(S_i \cap A = a)}{p(A = a)}$. If I am wrong, point out from the specific place where I went wrong.


  • $\begingroup$ Looks OK. In the displayed formula you intended $p(S_i\cap(A=a))$. The ultimate expression simplifies, and we can obtain the simplified version without introducing the $p$ and $1-p$. $\endgroup$ Mar 1, 2015 at 7:50
  • $\begingroup$ @AndreNicolas Yeah that union must be intersection I have corrected it. $\endgroup$
    – abkds
    Mar 1, 2015 at 7:52
  • 2
    $\begingroup$ Given that we had $a$ successes in $2m$ trials, and independence, the probability two specific trials resulted in success is $\frac{\binom{2m-2}{a-2}}{\binom{2m}{a}}$. Bu your calculation is good, leads to the same thing. $\endgroup$ Mar 1, 2015 at 7:57
  • 1
    $\begingroup$ Your method is sound, but I would be more careful with your notation: $S \in \{0, 1, \ldots, m\}$ is a random variable, and the way you have described your indicators, it seems that $S_i \in \{0, 1\}$ so these are also random variables. Consequently you should write $\Pr[S_i = 1 \mid A = a]$ and $\Pr[(S_i = 1) \cap (A = a)]$. $\endgroup$
    – heropup
    Mar 1, 2015 at 7:59
  • $\begingroup$ @heropup you are right. $\endgroup$
    – abkds
    Mar 1, 2015 at 8:01


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