Chapter 2, Problem 11SE

a.

To determine

Find $P\left(X\le 2\text{ and }Y\le 3\right)$.

Answer to Problem 11SE

The value of $P\left(X\le 2\text{ and }Y\le 3\right)$ is 0.3996.

Explanation of Solution

Given info:

The probability density function of X is,

$f\left(x\right)=\{\begin{array}{l}\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}x>0\text{ and }y>0\\ 0\text{ Otherwise}\end{array}$

Calculation:

The value of $P\left(X\le 2\text{ and }Y\le 3\right)$ is obtained as follows:

$\begin{array}{c}P\left(X\le 2\text{ and }Y\le 3\right)={\displaystyle {\int }_0^2{\displaystyle {\int }_0^3\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}}}dydx\\ =\frac{1}{6}{{\displaystyle {\int }_0^2\left[\frac{e^{-\frac{x}{2}-\frac{y}{3}}}{-\frac{1}{3}}\right]}}_0^{3}dx\\ =\frac{1}{2}{{\displaystyle {\int }_0^2\left[\left(-e^{-\frac{x}{2}-\frac{3}{3}}\right)+\left(e^{-\frac{x}{2}-\frac{0}{3}}\right)\right]}}_0^{3}dx\\ =\frac{1}{2}{\displaystyle {\int }_0^2\left[e^{-\frac{x}{2}}\left(1-e^{-1}\right)\right]}dx\end{array}$

$\begin{array}{c}=\frac{1}{2}\left(1-e^{-1}\right){\left(\frac{e^{-\frac{x}{2}}}{-\frac{1}{2}}\right)}_0^2\\ =\left(1-e^{-1}\right)\left(1-e^{-1}\right)\\ ={\left(1-e^{-1}\right)}^2\\ ={\left(0.6321\right)}^2\\ =0.3996\end{array}$

Thus, the value of $P\left(X\le 2\text{ and }Y\le 3\right)$ is 0.3996.

b.

To determine

Find the probability that both fuses last at least 3 years.

Answer to Problem 11SE

The probability that both fuses last at least 3 years is 0.0821.

Explanation of Solution

Calculation:

The probability that both fuses last at least 3 years represents the $P\left(X\ge 3\text{ and }Y\ge 3\right)$.

The value of $P\left(X\ge 3\text{ and }Y\ge 3\right)$ is obtained as follows:

$\begin{array}{c}P\left(X\ge 3\text{ and }Y\ge 3\right)={\displaystyle {\int }_3^{\infty }{\displaystyle {\int }_3^{\infty }\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}}}dydx\\ =\frac{1}{6}{{\displaystyle {\int }_0^{\infty }\left[\frac{e^{-\frac{x}{2}-\frac{y}{3}}}{-\frac{1}{3}}\right]}}_0^{\infty }dx\\ =\frac{1}{2}{{\displaystyle {\int }_0^{\infty }\left[\left(-e^{-\frac{x}{2}-\infty }\right)+\left(e^{-\frac{x}{2}-\frac{3}{3}}\right)\right]}}_0^{3}dx\\ =\frac{1}{2}{\displaystyle {\int }_0^{\infty }\left[e^{-\frac{x}{2}}\left(-0+e^{-1}\right)\right]}dx\end{array}$

$\begin{array}{c}=\frac{1}{2}\left(e^{-1}\right){\left(\frac{e^{-\frac{x}{2}}}{-\frac{1}{2}}\right)}_3^{\infty }\\ =e^{-1}\left(-0+e^{-\frac{3}{2}}\right)\\ =e^{-\frac{5}{2}}\\ =0.0821\end{array}$

Thus, the probability that both fuses last at least 3 years is 0.0821.

c.

To determine

Find the marginal probability density functions of X.

Answer to Problem 11SE

The marginal probability density functions of X is,

$f_X\left(x\right)=\{\begin{array}{l}\frac{1}{2}{e}^{-\frac{x}{2}}x>0\\ 0x\le 0\end{array}$

Explanation of Solution

Calculation:

The formula for finding marginal probability density function $f_X\left(x\right)$ is,

$f_X\left(x\right)={\displaystyle {\int }_{-\infty }^{\infty }f\left(x,y\right)dy}$

The marginal probability density function $f_X\left(x\right)$ is as follows:

For $x\le 0$:

Here, $x\le 0$ then $f\left(x,y\right)=0$.

Therefore,

$\begin{array}{c}{f}_X\left(x\right)={\displaystyle {\int }_{-\infty }^{0}0dy}\\ =\underset{\_}{0}\end{array}$

For $x>0$:

Here, $x>0$ then $f\left(x,y\right)=\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}$.

Therefore,

$\begin{array}{c}{f}_X\left(x\right)={\displaystyle {\int }_0^{\infty }\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}dy}\\ =\frac{1}{6}{e}^{-\frac{x}{2}}{\left[\frac{-e^{-\frac{y}{3}}}{\frac{1}{3}}\right]}_0^{\infty }\\ =\frac{1}{2}{e}^{-\frac{x}{2}}\left(-0+1\right)\\ =\frac{1}{2}{e}^{-\frac{x}{2}}\end{array}$

Thus, the marginal probability density functions of X is,

$f_X\left(x\right)=\{\begin{array}{l}\frac{1}{2}{e}^{-\frac{x}{2}}x>0\\ 0x\le 0\end{array}$

d.

To determine

Find the marginal probability density functions of Y.

Answer to Problem 11SE

The marginal probability density functions of Y is,

$f_Y\left(y\right)=\{\begin{array}{l}\frac{1}{3}{e}^{-\frac{y}{3}}y>0\\ 0y\le 0\end{array}$

Explanation of Solution

Calculation:

The formula for finding marginal probability density function $f_Y\left(y\right)$ is,

$f_Y\left(y\right)={\displaystyle {\int }_{-\infty }^{\infty }f\left(x,y\right)dx}$

The marginal probability density function $f_X\left(x\right)$ is as follows:

For $y\le 0$:

Here, $y\le 0$ then $f\left(x,y\right)=0$.

Therefore,

$\begin{array}{c}{f}_Y\left(y\right)={\displaystyle {\int }_{-\infty }^{0}0dy}\\ =\underset{\_}{0}\end{array}$

For $y>0$:

Here, $y>0$ then $f\left(x,y\right)=\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}$.

Therefore,

$\begin{array}{c}{f}_Y\left(y\right)={\displaystyle {\int }_0^{\infty }\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}dy}\\ =\frac{1}{6}{e}^{-\frac{y}{3}}{\left[\frac{-e^{-\frac{x}{2}}}{\frac{1}{2}}\right]}_0^{\infty }\\ =\frac{1}{3}{e}^{-\frac{y}{3}}\left(-0+1\right)\\ =\frac{1}{3}{e}^{-\frac{y}{3}}\end{array}$

Thus, the marginal probability density functions of Y is,

$f_Y\left(y\right)=\{\begin{array}{l}\frac{1}{3}{e}^{-\frac{y}{3}}y>0\\ 0y\le 0\end{array}$

e.

To determine

Explain whether X and Y are independent or not.

Answer to Problem 11SE

Yes, X and Y are not independent because $f\left(x,y\right)=f_X\left(x\right)f_Y\left(y\right)$.

Explanation of Solution

Calculation:

If $f_X\left(x\right)$ and $f_Y\left(y\right)$ are independent then $f\left(x,y\right)=f_X\left(x\right)f_Y\left(y\right)$.

Here,

$\begin{array}{c}f\left(x,y\right)=f_X\left(x\right)f_Y\left(y\right)\\ \frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}=\left(\frac{1}{2}{e}^{-\frac{x}{2}}\right)\left(\frac{1}{3}{e}^{-\frac{y}{3}}\right)\\ \frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}=\frac{1}{6}{e}^{-\frac{x}{2}-\frac{y}{3}}\end{array}$

Thus, X and Y are independent.