Variance question about fires in neighbourhoods Suppose that for any given year a neighbourhood will experience exactly zero fire with probability 0.1, exactly one fire with probability 0.3 and exactly two fires with probability 0.6. When there is a fire, the chance that the building on fire will have a structural damage is p. Assume all damage and fires are independent of one another. What is the variance for the number of buildings that will have structural damage due to fires in any given year?
I cant seem to get the answer. Would be delighted to be enlightened on how you solve this.
variance is v[y] = 1.5p-1.05$p^2$
 A: Let $Y$ be the number of buildings with structural damage due to fire. To find the variance of $Y$, we compute $E(Y^2)-(E(Y))^2$.
We can either find the distribution of $Y$, or use conditional expectations. We use the distribution approach, perhaps later doing it the other way.
Find first $\Pr(Y=0)$. This can happen in various ways: (i) $0$ fires; (ii) $1$ fire and no damage;(iii) $2$ fires and no damage. We actually don't really need $\Pr(Y=0)$, but for completeness it is $0.1+(1-p)(0.3)+(1-p)^2(0.6)$.
Similarly, $Y=1$ can happen in two ways: (i) $1$ fire and damage or (ii) $2$ fires but damage only once.
The probability that $Y=1$ is $p(0.3)+2p(1-p)(0.6)$.
Finally, $Y=2$ can only happen if there are two fires with damage each time. The probability is $p^2(0.6)$.
Now we need $E(Y)$ and $E(Y^2)$.
We have $E(Y)=(1)(0.3p+(0.6)2p(1-p))+(2)(0.6)p^2)=1.5p$.
Similarly, $E(Y^2)=(1^2)((0.3)p+(0.6)2p(1-p))+(2^2)(0.6p^2)=1.5p+1.2p^2$.
Now you have all the ingredients for computing the variance. It indeed turns out to be $1.5p-1.05p^2$.
