Discrete probability - Variance property
Given $E[Y]=2$ and $Var[Y]=3$, find $Var[3Y+4]$.
I used algebra and distributed the Var and obtained $$Var[3Y]+Var[4]$$
which would give $13$ but it makes no sense as to how the answer came to be $3^2 \times Var[Y]=27$.
Can someone explain this nonsense?
$\newcommand{\Var}{\operatorname{Var}}$ Actually $$\Var(aX+b) = a^2\Var(X).$$
We can verify via one of the variance formulas \begin{align*} \Var(aX+b) &= E[(aX+b)^2]-(E[aX+b])^2\\ &= E[a^2X^2+2abX+b^2]-(E[aX]+E[b])^2\\ &=a^2E[X^2]+2abE[X]+E[b^2]-(aE[X]+b)^2\\ &=a^2E[X^2]+2abE[X]+b^2-(a^2(E[X])^2+2abE[X]+b^2)\\ &=a^2(E[X^2]-(E[X])^2)\\ &= a^2\Var(X) \end{align*} wherein throughout I used the fact that $E[aX] = aE[X]$ and $E[b] = b$.
So in OP $$\Var(3Y+4) = 3^2\Var(Y) = 9\cdot 3 = 27.$$
You can find more properties in the Wikipedia article on variance.