Variance and expected values

Question

i have the following question:

A coin is flipped n times with the probability of $\frac35$ for heads. Each heads gets you 2 points and tails -3 points. Let $R$ be the total sum of points after n flips. Calculate the Expected value and the variance of $R$

My question is this: can i define $X$-"number of heads" and $N$-"number of tosses" and say that $R = 2X-3(N-X)$, and go on from there according to the formulas, assigning $N=n$ wherever required? Or do i have to define $Y$-"number of tails"? because the latter is difficult when i have to calculate $E(XY)$ which i have no clue how to.

Answer 1

Representing the number of tails with N - X is perfectly fine and is recommended for this situation. If you set X to the number of heads and Y to the number of tails, it still follows that X + Y = N, the total number of tosses. All you're doing in this cases is rewriting the above equation to isolate Y to equal N - X, which is an acceptable substitution.

Answer 2

You don't need to define the random variable $Y$, as you can write the number of tails as a function of $n$ and $X$.

You deduced $R=2X-3n+3X=5X-3n$ in the correct way. The expectation is then given by:

$E[R]=E[5X-3n]=5E[X]-3n$

$X$ follows a binomial distribution with parameters $(3/5, n)$, so $E[X]=3n/5$. It follows that $E[R]=0$. Moreover the variance of $X$ is given by $ \frac{3}{5}n(1-\frac{3}{5})=\frac{6n}{25}$ .

Therefore $Var(R)=Var(5X-3n)=Var (5X)=25 Var(X)=6n$