Question

A web-based store is stuck with a very large set of three kinds of unsold Christmas snow globes: Globes with a Nutcracker figurine, globes with a Santa figurine, and globes with a Snowman (see pictures below from left to right). 2.o Fifty percent of the snow globes have the Nutcracker, 20% have Santa, and 30% have the Snowman. The store runs a special where they offer 10 randomly chosen snow globes for $100-a roughly "10 for the price of 4" special. (3 pts) Define random variables for each of the types of snow globes and give the name and parameters of the joint distribution of these random variables. a) b) (3 pts) Find the probability of getting 3 Nutcracker snow globes and 7 Snowman snow globes in a set of 10 snow globes. Show all work. (6 pts) Let X= three times the number of nutcrackers plus two times the number of snowmen. the mean and variance of X c) Find

Answer 1

a) Out of n=10 randomly chosen snow globes, let

• $\begin{align*}X_1\end{align*}$ indicate the number of Nutcracker snow globes
• $\begin{align*}X_2\end{align*}$ indicate the number of Santa snow globes
• $\begin{align*}X_3\end{align*}$ indicate the number of Snowman snow globes

The joint distribution of $\begin{align*}X_1,X_2,X_3\end{align*}$ has a multinomial distribution with parameters n=10 trials and probabilities $p_1=0.50,p_2=0.20,p_3=0.30$

the joint pmf is given by

$\begin{align*} P(x_1,x_2,x_3)&=\frac{n!}{x_1!x_2!x_3!}p_1^{x_1}\times p_2^{x_2}\times p_3^{x_3}\\ &=\frac{10!}{x_1!x_2!x_3!}0.50^{x_1}\times 0.20^{x_2}\times 0.30^{x_3}\\ \end{align*}$

Formally the joint pmf  of $\begin{align*}X_1,X_2,X_3\end{align*}$ is

$\begin{align*} P(x_1,x_2,x_3)&=\frac{10!}{x_1!x_2!x_3!}0.50^{x_1}\times 0.20^{x_2}\times 0.30^{x_3},\quad x_1,x_2,x_3= 0,1,2,...,10; x_1+x_2+x_3=10\\ \end{align*}$

b) The probability of getting 3 Nutcracker snow globes, 0 Santa snow globes and 7 Snowman snow globes in a set of 10 snow globes is

$\begin{align*} P(3,0,7)&=\frac{10!}{3!0!7!}0.50^3\times 0.20^0\times 0.30^7\\ &=\frac{10\times 9\times ...\times 1}{(3\times 2\times 1)\times (1)\times (7\times 6\times ...\times 1)}0.50^3\times 1\times 0.30^7\\ &=\frac{10\times 9\times 8}{(3\times 2\times 1)}0.50^3\times 1\times 0.30^7\\ &=0.0032805 \end{align*}$

The probability of getting 3 Nutcracker snow globes, and 7 Snowman snow globes in a set of 10 snow globes is 0.0033

c) Let $\begin{align*}X_1\end{align*}$ indicate the number of Nutcracker snow globes out of 10 randomly selected globes. The marginal pmf of $\begin{align*}X_1\end{align*}$ is a Binomial distribution with parameters number of trails, n=10 and probability of success p=0.50

The expected value of $\begin{align*}X_1\end{align*}$ is (using the result for Binomial distribution)

$\begin{align*} E(X_1)=n\times p=10\times 0.50=5 \end{align*}$

The variance of $\begin{align*}X_1\end{align*}$ is

$\begin{align*} Var(X_1)=n\times p\times (1-p)=10\times 0.50\times (1-0.50)=2.5 \end{align*}$

Let $\begin{align*}X_3\end{align*}$ indicate the number of Snowman snow globes out of 10 randomly selected globes. The marginal pmf of $\begin{align*}X_3\end{align*}$ is a Binomial distribution with parameters number of trails, n=10 and probability of success p=0.30

The expected value of $\begin{align*}X_3\end{align*}$ is (using the result for Binomial distribution)

$\begin{align*} E(X_3)=n\times p=10\times 0.30=3 \end{align*}$

The variance of $\begin{align*}X_3\end{align*}$ is

$\begin{align*} Var(X_3)=n\times p\times (1-p)=10\times 0.30\times (1-0.30)=2.1 \end{align*}$

The covariance of $\begin{align*}X_1,X_3\end{align*}$ is (using the standard result of a multinomial distribution)

$\begin{align*} Cov(X_1,X_3)=-np_1p_3=-10\times 0.5\times 0.3=-1.5\end{align*}$

Now $\begin{align*} X=3X_1+2X_3\end{align*}$

For any 2 random variables Y, Z and constants a,b we have the following results

$\begin{align*} E(aY+bZ)=aE(Y)+bE(Z)\end{align*}$

$\begin{align*} Var(aY+bZ)=a^2Var(Y)+b^2Var(Z)+2abCov(Y,Z)\end{align*}$

The expected value of X is

$\begin{align*} E(X)&=E(3X_1+2X_3)=3E(X_1)+2E(X_3)=3\times 5+2\times 3 = 21\end{align*}$

the variance of X is

$\begin{align*} Var(X)&=Var(3X_1+2X_3)\\ &=3^2Var(X_1)+2^2Var(X_3)+2\times 3\times 2\times Cov(X_1,X_3)\\ &=3^2\times 2.5+2^2\times 2.1+2\times 3\times 2\times (-1.5)\\ &=12.9 \end{align*}$