Online Dating

Hal has a $(1/3)\times (1/10)=1/30$ chance of messaging any given person. If he has sent $100$ messages, then on average, he will have viewed $30\times 100=3000$ people. Of these people, on average, $3000/10=300$ of them will be interesting, and $3000/3=1000$ of them will be attractive. Of the average $300$ interesting people, Hal will have messaged $100$ , leaving $300-100=\boxed{200}$ interesting people dismissed.

I would like to add this answer to the neat one Miles wrote.

Let $A$ and $I$ be the two events that the next person Hal checks out is attractive and interesting respectively, and let $N$ be the number of interesting people Hal passes up before his message number $h = 100$ to both-attractive-and-interesting people. Let's condition $N$ on the outcome of $M$ , the number of people Hal eventually checks out, with $M$ taking possible values in $[h, \infty)$ .

Person number $M$ triggers the final message and ends the process, so Hal sends messages to $h - 1$ of $M - 1$ previous persons, of which $M - h$ are likely to be passed up. We know for sure they are not in the intersection $A \cap I$ . This makes the conditioned variate $N \, | \, M$ a binomial variate with parameters $n = M - h$ and $p$ :

$p = \mathbb{P}\{ I \cap \overline{A} \, | \, \overline{ I \cap A} \} = \frac{ \mathbb{P}\{ I \cap \overline{A}, \overline{ I \cap A} \} }{ \mathbb{P}\{ \overline{ I \cap A} \}} = \frac{ \mathbb{P}\{ I \cap \overline{A} \} }{ \mathbb{P}\{ \overline{ I \cap A} \}} = \frac{i - i \, a}{1 - i \, a}$ $\mathbb{E}[N] = \mathbb{E}[\mathbb{E}[N \, | \, M]] = \mathbb{E}[n \, p] = h \, (1 / a - 1)$

Interestingly, the expected number of merely interesting people passed up is independent of the chance of finding a person interesting $i$ , with $0 < i < 1$ .