Question

The joint probability mass function of X and Y is given by

$$\begin{gathered} p(1,1)=\frac{1}{8}p(1,2)=\frac{1}{4} \\ p(2,1)=\frac{1}{8}p(2,2)=\frac{1}{2} \end{gathered}$$

Short answer

(a) The conditional mass function of X:

$$\begin{gathered} ForY=1, \\ P(X=1\mid Y=1)=\frac{1}{2} \\ P(X=2\mid Y=1)=\frac{1}{2} \\ ForY=2, \\ P(X=1\mid Y=2)=\frac{1}{3} \\ P(X=2\mid Y=2)=\frac{2}{3} \end{gathered}$$

(b) X and Y are not independent

(c) Corresponding probabilities,

$$\begin{gathered} ForXY\le 3: \\ P(XY\le 3)=\frac{1}{2} \\ ForX+Y>2: \\ P(X+Y>2)=\frac{7}{8} \\ ForX/Y>1: \\ P\left(X/Y>1\right)=\frac{1}{8} \end{gathered}$$

Step-by-step solution

Given information (part a)

Y is represented by

$p(1,1)=\frac{1}{8},p(1,2)=\frac{1}{4},p(2,1)=\frac{1}{8},p(2,2)=\frac{1}{2}$

also, Y = i for$i\in \left\{1,2\right\}$

Explanation (part a)

Probability for Y$=1$,

$P(Y=1)=p(1,1)+p(2,1)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$

Therefore, the conditional mass function of X:

For $Y=1$

localid="1647238313124" $P(X=1\mid Y=1)=\frac{P(X=1,Y=1)}{P(Y=1)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}$

and

$P(X=2\mid Y=1)=\frac{P(X=2,Y=1)}{P(Y=1)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}$

for $Y=2$

$P(X=1\mid Y=2)=\frac{P(X=1,Y=2)}{P(Y=2)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

and

$P(X=2\mid Y=2)=\frac{P(X=2,Y=2)}{P(Y=2)}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$

Given information (part b) 

Y is represented by

$p(1,1)=\frac{1}{8},p(1,2)=\frac{1}{4},p(2,1)=\frac{1}{8},p(2,2)=\frac{1}{2}$

Explanation (part b)

Form part (a) we have

$P(X=1\mid Y=1)=\frac{P(X=1,Y=1)}{P(Y=1)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}$

then

$P(X=1)=p(1,1)+p(1,2)=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$

thus

$P(X=1\mid Y=1)\ne P(X=1)$

Therefore X and Y are not independent.

Given information (part c)

Y is represented by

$p(1,1)=\frac{1}{8},p(1,2)=\frac{1}{4},p(2,1)=\frac{1}{8},p(2,2)=\frac{1}{2}$

Explanation (part c)

Corresponding probability is

Probability for $XY\le 3:$

localid="1647238518761" $P(XY\le 3)=p(1,1)+p(1,2)+p(2,1)=\frac{1}{8}+\frac{1}{4}+\frac{1}{8}=\frac{1}{2}$

Probabilities for $X+Y>2$:

$P(X+Y>2)=p(1,2)+p(2,1)+p(2,2)=\frac{1}{4}+\frac{1}{8}+\frac{1}{2}=\frac{7}{8}$

Probabilities for $X/Y>1$:

$P\left(X/Y>1\right)=p\left(2,1\right)=\frac{1}{8}$