Compute $\ Var(X+Y+Z) $ where $\ X,Y,Z \sim Binomial $

Question

Suppose I throw 3 fair dice 30 times.

Let,

X = no' of throws in which we don't get 4
Y = no' of throws in which we get 4 in only one die (out of 3)
Z = no' of throws in which we get 4 in exactly two dice (out of 3)

Compute $\ Var(X+Y+Z) $

by definition $\ Var(X+Y+Z) = E[(X+Y+Z)^2] + (E[X+Y+Z])^2 $

$\ E[X+Y+Z] = E[X] + E[Y] + E[Z] = 29.8611$ because each variable of a multinomial distribution is of binomial distribution. so I know $\ (E[X+Y+Z])^2 = 29.8611^2 $ but how do I calculate $\ E[(X+Y+Z)^2] $ ?

Suggestions please?

Answer 1

These $X$, $Y$, and $Z$ are strongly related to each other - to the point where we'd rather not work with them directly in calculating the variance. In particular, $X+Y+Z+W=30$, where $W$ is the number of throws in which all three dice come up fours. The variance of $X+Y+Z$ is the same as that of $W$. Can you calculate that?

Answer 2

Using the notation from @jmerry, let $W$ be the number of tosses in which all 3 dice show 4. Thus, $X+Y+Z+W=30$, i.e. $X+Y+Z=30-W$ implying $Var(X+Y+Z)=Var(30-W)$.

Let's consider $Q=30-W$.

$\mathbb{E}[Q] = 30-\mathbb{E}[W]$

$\mathbb{E}[Q^2] = 900 + \mathbb{E}[W^2] - 60\mathbb{E}[W]$

$Var[Q] = \mathbb{E}[Q^2] - \mathbb{E}[Q]^2$

$=900 + \mathbb{E}[W^2] - 60\mathbb{E}[W] - (30-\mathbb{E}[W])^2$

$=\mathbb{E}[W^2] -\mathbb{E}[W]^2$

$=Var(W)$ (by definition)

Thus, $Var(30-W) = Var(W) = Var(X+Y+Z)$.

$W$ follows a binomial distribution with parameters $N=30$ and $p = \left( \frac{1}{6} \right)^3$. You can calculate the variance as $Np(1-p)$.