Suppose that $75$% of the population is vaccinated against a certain disease $D$. The probability that a vaccinated person has to be hospitalised following the disease is equal to $0.05$, while for a non-vacinated it is ten times greater. Let's denote by $X$ a number of people hospitalized who have the disease (at a given day), and by $Y$ a number of vaccinated people among them. Compute conditional expectance $\mathbb{E}[Y \mid X = 50]$.
Obviously $\mathbb{E}[Y \mid X = 50]= \sum\limits_y y\frac{\mathbb{P}(Y, X = 50)}{\mathbb{P}(X = 50)}$ but I struggle to take into account the fact that both $X$ and $Y$ are counts.
Obviuosly $\mathbb{P}(D)= 0.75, \ \mathbb{P}(D^c)=0.25, \ \mathbb{P}(H \mid V, D)=0.05, \ \mathbb{P}(H \mid V^c, D) = 0.5$, where
$D$ - a pearson has a disease
$V$ - a pearson is vaccinated
$H$ - a pearson is hospitalized.
I'd be glad for any help or tip.
Let $A$ denote the probability that a hospitalized person was vaccinated.
Then, $E[Y|X=50] = 50 \times A,$ since the event of one person being hospitalized (vaccinated or not) is presumably independent of any other people being hospitalized, vaccinated or not.
$$A = \frac{p\left(\text{Vaccinated, Hospitalized}\right)}{p\left(\text{Hospitalized}\right)}$$
$$= \frac{0.75 \times 0.05}{[ ~0.75 \times 0.05 ~] + [ ~0.25 \times 0.5 ~]} = \frac{0.0375}{0.0375 + 0.125} \approx 0.23076923.$$
Therefore,
$$E[Y|X=50] = 50 \times A = 50 \times 0.23076923 \approx 11.53846154.$$