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While Fred is sleeping one night, $X$ legitimate emails and $Y$ spam emails are sent tohim. Suppose that X and Y are independent, with $X$ ~$ Pois(10)$ and $Y $ ~ $Pois(40)$.When he wakes up, he observes that he has $30$ new emails in his inbox. Given this information, what is the expected value of how many new legitimate emails he has?

Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.

Due to independence, $p(X,Y)=p(X)p(Y)$.

As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.

We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:

$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.

The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.

Finally, your result is given by:

$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.

Now, it's just a matter of computing $k$, $f(X)$ and the sum above.