Chapter 5.4, Problem 68E

a.

To determine

Find the joint probability distribution of $Y_1$ and $Y_2$.

Answer to Problem 68E

The joint probability distribution of $Y_1$ and $Y_2$ is,

$p\left(y_1,y_2\right)=\left(\begin{array}{l}2\\ y_1\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{2-y_1}{\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2},y_1=0,1,2\text{and}{y}_2=0,1$.

Explanation of Solution

Calculation:

Consider that $Y_1$ and $Y_2$ are two discrete real valued random variables. Then the joint probability mass function is defined as $p\left(y_1,y_2\right)$. In addition, the marginal probability functions of $Y_1$ and $Y_2$ are $p_1\left(y_1\right)$ and $p_2\left(y_2\right)$, respectively.

The two discrete random variables $Y_1$ and $Y_2$ are said to be independent if and only if $p\left(y_1,y_2\right)=p_1\left(y_1\right)p_2\left(y_2\right)$, where $p\left(y_1,y_2\right)$ is the joint probability mass function of $Y_1$ and $Y_2$ and $p_1\left(y_1\right),p_2\left(y_2\right)$ are the marginal probability mass functions of $Y_1$ and $Y_2$, respectively.

Binomial distribution:

A random variable Y is said to follows binomial distribution if and only if the probability mass function is defined as,

$p\left(y\right)=\left(\begin{array}{l}n\\ y\end{array}\right)p^y{\left(1-p\right)}^{n-y},y=0,1,2,\mathrm{...},n\text{and}0\le p\le 1$, where n is the total number of trials and p is the probability of success.

According to the question $Y_1$ take values as 0(no customer), 1(one customer), 2(two customers) and $Y_2$ take values as 0(no customer), 1(one customer).

Hence, according to the given question the probability mass function of $Y_1$ is,

$p_1\left(y_1\right)=\left(\begin{array}{l}2\\ y_1\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{2-y_1},y_1=0,1,2.$

Similarly, according to the given question the probability mass function of $Y_2$ is,

$p_2\left(y_2\right)=\left(\begin{array}{l}2\\ y_2\end{array}\right){\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2},y_2=0,1.$

As $Y_2$ takes only two values, such that, 0 and 1, then the probability distribution function of $Y_2$ can be written as Bernoulli distribution function.

That is,

$p_2\left(y_2\right)={\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2},y_2=0,1.$

As these two random variables are independent then the joint probability distribution would be the product of the marginal distributions.

That is,

$\begin{array}{c}p\left(y_1,y_2\right)=\left(\begin{array}{l}2\\ y_1\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{2-y_1}\left(\begin{array}{l}2\\ y_2\end{array}\right){\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2}\\ =\left(\begin{array}{l}2\\ y_1\end{array}\right)\left(\begin{array}{l}2\\ y_2\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{2-y_1}{\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2}\end{array}$

Thus, the joint probability distribution of $Y_1$ and $Y_2$ $p\left(y_1,y_2\right)=\left(\begin{array}{l}2\\ y_1\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{2-y_1}{\left(0.3\right)}^{y_2}{\left(1-0.3\right)}^{1-y_2},y_1=0,1,2\text{and}{y}_2=0,1$.

b.

To determine

Find the probability that not more than one of the three customers will spend more than $50.

Answer to Problem 68E

The probability that not more than one of the three customers will spend more than $50 is 0.864.

Explanation of Solution

Calculation:

The probability that not more than one of the three customers will spend more than $50 can be written as $P\left(Y_1+Y_2\le 1\right)$.

Hence, using the joint probability function the required probability is obtained as,

$P\left(Y_1+Y_2\le 1\right)=p\left(0,0\right)+p\left(0,1\right)+p\left(1,0\right)$.

The value of $p\left(0,0\right)$ is obtained as,

$\begin{array}{c}p\left(0,0\right)=\left(\begin{array}{l}2\\ 0\end{array}\right){\left(0.2\right)}^0{\left(1-0.2\right)}^{2-0}{\left(0.3\right)}^0{\left(1-0.3\right)}^{1-0}\\ =\left(1\right)\left(1\right)\left(0.64\right)\left(1\right)\left(0.7\right)\\ =0.448\end{array}$

The value of $p\left(1,0\right)$ is obtained as,

$\begin{array}{c}p\left(1,0\right)=\left(\begin{array}{l}2\\ 1\end{array}\right){\left(0.2\right)}^1{\left(1-0.2\right)}^{2-1}{\left(0.3\right)}^0{\left(1-0.3\right)}^{1-0}\\ =\left(2\right)\left(0.2\right)\left(0.8\right)\left(1\right)\left(0.7\right)\\ =0.224\end{array}$

The value of $p\left(0,1\right)$ is obtained as,

$\begin{array}{c}p\left(0,1\right)=\left(\begin{array}{l}2\\ 0\end{array}\right){\left(0.2\right)}^0{\left(1-0.2\right)}^{2-0}{\left(0.3\right)}^1{\left(1-0.3\right)}^{1-1}\\ =\left(1\right)\left(1\right)\left(0.64\right)\left(0.3\right)\left(1\right)\\ =0.192\end{array}$

Thus,

$\begin{array}{c}P\left(Y_1+Y_2\le 1\right)=0.448+0.224+0.192\\ =0.864\end{array}$

Thus, the probability that not more than one of the three customers will spend more than $50 is 0.864.