91Ó°ÊÓ

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$i∈\left\{1,2\right\}$

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

and

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

for $Y=2$

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

and

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

Y is represented by

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

Form part (a) we have

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

then

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

thus

$P(X=1âˆ\pounds Y=1)≠P(X=1)$

Therefore X and Y are not independent.

Y is represented by

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

Corresponding probability is

Probability for $XY≤3:$

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

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 $X/Y>1$:

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