Calls that become sales

Idea. Splitting a Poisson process at random gives two independent Poisson processes. So knowing the number of sales tells you nothing about the number of non-sales.

Step 1: thinning. Let $N$ be the number of calls in the hour, $S$ the sales and $F=N-S$ the other calls. Then $S\sim \text{Poisson}(6\cdot \frac{1}{3})=\text{Poisson}(2)$ and $F\sim \text{Poisson}(6\cdot \frac{2}{3})=\text{Poisson}(4)$, and $S$ and $F$ are independent.

Step 2: condition. $$\begin{array}{rl}\mathbb{E}[N\mid S=4]& =4+\mathbb{E}[F\mid S=4]\\ & =4+\mathbb{E}[F]=4+4=8.\end{array}$$

Why independence is surprising. One might expect $4$ sales to suggest about $4\cdot 3=12$ calls. That reasoning would be right if the total number of calls were fixed in advance. Here it is random, and with a Poisson count the two kinds of calls carry no information about each other.

Check without the trick. $\mathbb{P}(F=f\mid S=4)$ can be computed from $\mathbb{P}(N=4+f)\left(\genfrac{}{}{0ex}{}{4+f}{4}\right){\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^f$; after simplification it is exactly the $\text{Poisson}(4)$ probability of $f$.