Expected value and variance of random variable of bus 
The bus company has $m$ bus. The bus had two side mirrors. In the streets of Munich, each of the $2m$ mirrors breaks oﬀ independently with probability $p$. Let $n$ be the total number of broken mirrors, $L$ be the number of cars with a broken left mirror, $R$ be the number of cars with a broken right mirror, and $x$ be the number of cars with no mirrors.

*

*The expected value of $L$ and variance $R$.


*The expected value and variance of $n$.


*The expected value and variance of $x$.


*The expected value and variance of $L-R$.

I have no idea with the remaining and I am not really sure is my answer is logical.

*

*This should be a binominal distribution so the expected value should be $Lp$ and variance is $L^2p-(Lp)^2$.


*Expected value is $np$ and the variance is $np(1-p)$.
 A: Hints (on how to compute the expected values):
In both of your answers, the expected value of your variable depends on the variable itself, which is not satisfactory.


*

*For each of the $m$ buses, flip one coin that comes up heads with probability $p$. If you get a heads, the left mirror is broken. How many do you expect to be broken after you've flipped one coin for each bus?

*The $2m$ mirrors break off with probability $p$, so we'd expect there to be $E(n)=\dots$ broken mirrors. 

*Now flip two coins for each bus. If both comes up heads, both the mirrors are broken. What is the probability of this? 

*Use linearity of expectation.
For the variance part, have you tried applying the definition of the variance?
A: In cases 1,2,3 you are dealing with binomial distribution.


*

*1) Parameters $m$ and $p$

*2) Parameters $2m$ and $p$

*3) Parameters $m$ and $p^2$





*

*4)


$L$ and $R$ have the same binomial distribution. Moreover $L$ and $R$ are independent as well (the breaking off of a window at one side has no influence on the breaking off of a window at the other side).
Can you find the expectation and variance of $L-R$ yourself? Remember that the variance of the sum of two independent variables equalizes the sum of the involved variances.
