Suppose that a transport company operates $40$ trams and $80$ buses. Every day, independently of each other, each tram breaks down with probability $0.01$, and each bus breaks down with probability $0.02$ (independently of each other and of the trams). Let $X$ be the number of trams breaking down during a day, and $Y$ be the number of buses breaking down on the same day. Calculate the covariance of $X$ and $X · Y$.
My solution: $$X: n = 40, p = 0.01, M(X) = np = 0.4$$ $$Y: n = 80, p = 0.02, M(Y) = np = 1.6$$ $$M(XY) = M(X)M(Y) = 0.64$$ $$cov(X, XY)=M(XXY) - M(X)M(XY)$$ And I dont know how to calculate $M(XXY)$. Because the idea of $M(XXY) = M(X)M(XY)$ leads to $cov(X,XY) = 0$
I presume that $M$ in your solution means Expectation.
In this problem, $X\sim \text{Bin}(40, 0.01)$ and $Y\sim \text{Bin}(80, 0.02)$ where $X$ and $Y$ are independent. In particular $$ \text{Cov}(X, XY)=EX^2Y-EXEXY $$ To compute $EX^2Y$ we use the fact that if $X, Y$ are independent then so are $\varphi(X)$ and $\psi (Y)$ where $\varphi$ and $\psi$ are any (measurable) functions. In particular $X^2$ and $Y$ are independent and so $$ EX^2Y=EX^2EY $$ which you can compute.