Q.6.5 - A First Course in Probability

Question

Suppose that X, Y, and Z are independent random variables that are each equally likely to be either 1 or 2. Find the probability mass function of 

(a) $XYZ$, 

(b) $XY + XZ + YZ$, and

(c) $X^2+YZ$

Step-by-Step Solution

Answer

a) The probability of the mass function $XYZ$:

$\begin{array}{r}\text{ At }j=1:p_1=\frac{1}{8}\\ \text{ At }j=2:p_2=\frac{3}{8}\\ \text{ At }j=4:p_4=\frac{3}{8}\\ \text{ At }j=8:p_8=\frac{1}{8}\end{array}$

b) The probability of the mass function $XY+YZ+ZX$:

$\begin{array}{r}\text{ At }j=3:p_3=\frac{1}{8}\\ \text{ At }j=5:p_5=\frac{3}{8}\\ \text{ At }j=8:p_8=\frac{3}{8}\\ \text{ At }j=12:p_{12}=\frac{1}{8}\end{array}$

c) The probability of the mass function $X^2+YZ$

$\begin{array}{r}\text{ At }j=2:p_2=\frac{1}{8}\\ \text{ At }j=3:p_3=\frac{1}{4}\\ \text{ At }j=5:p_5=\frac{1}{4}\\ \text{ At }j=6:p_6=\frac{1}{4}\\ \text{ At }j=8:p_8=\frac{1}{8}\end{array}$

Part (a) - Step 1: To find

The probability of the mass function $XYZ$

Part (a) - Step 2: Explanation

Given : Independent  random variables are $XYZ$

Consider

$\begin{array}{rl}& \begin{array}{rl}{p}_j& =P\{XYZ=j\}\\ \text{If}j=1,X=Y=Z=1\text{then:}& \\ p_1& =P\{XYZ=1\}\\ p_1& =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ p_1& =\frac{1}{8}\\ \text{If}j=2& (X=2,Y=1,Z=1;X=1,Y=2,Z=1;X=1,Y=1,Z=2)\text{then:}\\ p_2& =P\{XYZ=2\}\\ p_2& =(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})\\ p_2& =\frac{3}{8}\\ \text{If}j=4,& (X=2,Y=2,Z=1;X=2,Y=1,Z=2;X=1,Y=2,Z=2)\text{then:}\\ p_4& =P\{XYZ=4\}\\ p_4& =(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})\\ p_4& =\frac{3}{8}\\ \text{If}j=8,& (X=2,Y=2,Z=2)\text{then:}\\ p_8& =P\{XYZ=8\}\\ p_8& =(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})\\ p_8& =\frac{1}{8}\end{array}\end{array}$

Part (b) - Step 3: To find

The probability of the mass function $XY+YZ+ZX$ is:

Part (b) - Step 4:Explanation

Consider

$\begin{array}{r}\begin{array}{c}{p}_j=P\{XY+XZ+YZ=j\}\\ \text{ If }j=3,X=Y=Z=1\text{ then: }\\ p_3=P\{XY+XZ+YZ=3\}\\ p_3=\frac{1}{8}\\ \text{ If }j=5,(X=2,Y=1,Z=1;X=1,Y=2,Z=1;X=1,Y=1,Z=2)\text{ then: }\\ p_5=P\{XY+XZ+YZ=5\}\\ p_5=\frac{3}{8}\\ \text{ If }j=8,(X=2,Y=2,Z=1;X=2,Y=1,Z=2;X=1,Y=2,Z=2)\text{ then: }\\ p_8=P\{XY+XZ+YZ=8\}\\ p_4=\frac{3}{8}\\ \text{ If }j=12,(X=2,Y=2,Z=2)\text{ then: }\\ p_{12}=P\{XY+XZ+YZ=12\}\\ p_{12}=\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)\\ p_{12}=\frac{1}{8}\end{array}\end{array}$

Part (c) - Step 5: To find

The probability mass function of  $X^2+YZ$

Part(c) - Step 6: Explanation

To given: Independent random variables $X,Y,Z$

Consider

$\begin{array}{r}{p}_j=P\left\{X^2+YZ=j\right\}\\ \text{ If }j=2,X=Y=Z=1\text{ then: }\\ p_2=P\left\{X^2+YZ=2\right\}\\ p_2=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ p_2=\frac{1}{8}\\ \text{ If }j=3,(X=1,Y=2,Z=1;X=1,Y=1,Z=2)\text{ then: }\\ p_3=P\left\{X^2+YZ=3\right\}\\ p_3=\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)\\ p_3=\frac{1}{4}\\ \text{ If }j=5,(X=2,Y=1,Z=1;X=1,Y=2,Z=2)\text{ then: }\\ p_5=P\left\{X^2+YZ=5\right\}\\ p_5=\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)\\ p_5=\frac{1}{4}\\ \text{ If }j=6,(X=2,Y=1,Z=1;X=2,Y=2,Z=1)\text{ then: }\\ p_6=P\left\{X^2+YZ=6\right\}\\ p_6=\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)\\ p_6=\frac{1}{4}\\ \text{ If }j=8,(X=2,Y=2,Z=2)\text{ then: }\\ p_8=\left(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right)\\ p_0=\frac{1}{8}\end{array}$