91Ó°ÊÓ

a.

Define${\mathrm{N}}_1$and${\mathrm{N}}_2$as random variables that represent the numbers rolled on the first and second dice, respectively.

We know that ${\mathrm{N}}_1$and ${\mathrm{N}}_2$are unrelated, and thatlocalid="1649438807914" ${\mathrm{N}}_1,{\mathrm{N}}_2~\mathrm{DUnif}(1,.....,6)$.

localid="1649438811556" $\mathrm{X}=\max ({\mathrm{N}}_1,{\mathrm{N}}_2)$andlocalid="1649438816033" $\mathrm{Y}={\mathrm{N}}_1+{\mathrm{N}}_2$are the values here.

As a result,localid="1649438836153" $\mathrm{X}∈\{1,.......,6\}$andlocalid="1649438842350" $\mathrm{Y}∈\{2,........,12\}$.

Also, we very surely have that localid="1649438795160" $\mathrm{X}<\mathrm{Y}$.

Take any localid="1649438830987" $\mathrm{k}<\mathrm{l}$,where localid="1649438847895" $\mathrm{k}$andlocalid="1649438853351" $\mathrm{l}$are from the above-mentioned ranges.

Consider the localid="1649438859116" $\mathrm{X}=\mathrm{k}$,localid="1649438864665" $\mathrm{Y}=\mathrm{l}$occurrence.

This means that any die's highest value is localid="1649438869853" $\mathrm{k}$,and the sum of both dice islocalid="1649438876355" $\mathrm{l}$.

Note that the only conceivable pairs of localid="1649438881595" $({\mathrm{N}}_1,{\mathrm{N}}_2)$are localid="1649438891363" $(\mathrm{k},\mathrm{l}-\mathrm{k})$andlocalid="1649438886762" $(\mathrm{l}-\mathrm{k},\mathrm{k})$

for localid="1649438907720" $\mathrm{l}<2\mathrm{k}$

localid="1649438902743" $(\mathrm{k},\mathrm{k})$is possible forlocalid="1649438897219" $\mathrm{l}=2\mathrm{k}$

As a result, the needed PMF is

$\mathrm{P}(\mathrm{X}=\mathrm{k},\mathrm{Y}=\mathrm{l})=\left\{\begin{array}{l}\frac{2}{36},\mathrm{K}<\mathrm{l}<2\mathrm{K}\\ \frac{1}{36},\mathrm{l}=2\mathrm{K}\end{array}\right.$

b.

Here, $\mathrm{X}={\mathrm{N}}_1$and$\mathrm{Y}=\max ({\mathrm{N}}_1,{\mathrm{N}}_2)$are the values.

Both variables are in thelocalid="1649438741184" $\{1,......,6\}$range.

$\mathrm{P}(\mathrm{X}=\mathrm{k},\mathrm{Y}=\mathrm{l})=\mathrm{P}(\mathrm{Y}=\mathrm{l}âˆ\pounds \mathrm{X}=\mathrm{k})\mathrm{P}(\mathrm{X}=\mathrm{k})$

Assume thatlocalid="1649438745541" $\mathrm{k}=\mathrm{l}$and that localid="1649438749086" $\mathrm{X}=\mathrm{k}$is handed to us. In that situation,localid="1649438756528" ${\mathrm{N}}_2$can be any number betweenlocalid="1649438762431" $1,.......,\mathrm{k}$to get the requisitelocalid="1649438732063" $\mathrm{Y}=\mathrm{l}$.

Hence

$\mathrm{P}(\mathrm{Y}=\mathrm{l}âˆ\pounds \mathrm{X}=\mathrm{k})\mathrm{P}(\mathrm{X}=\mathrm{k})=\frac{\mathrm{k}}{6}·\frac{1}{6}$

$=\frac{\mathrm{k}}{36}$

Iflocalid="1649438769860" $\mathrm{k}<\mathrm{l}$and we are given that localid="1649438776197" $\mathrm{X}=\mathrm{k},{\mathrm{N}}_2$must be equal tolocalid="1649438782352" $\mathrm{l}$to obtainlocalid="1649438787488" $\mathrm{Y}=\mathrm{l}$.

So, in that case

$\mathrm{P}(\mathrm{Y}=\mathrm{l}âˆ\pounds \mathrm{X}=\mathrm{k})\mathrm{P}(\mathrm{X}=\mathrm{k})=\frac{1}{6}·\frac{1}{6}$

$=\frac{1}{36}$

c.

$\mathrm{X}=\min ({\mathrm{N}}_1,{\mathrm{N}}_2)$andlocalid="1649438694059" $\mathrm{Y}=\max ({\mathrm{N}}_1,{\mathrm{N}}_2)$are the values here.

We almost likely havelocalid="1649438699750" $\mathrm{X}≤\mathrm{Y}$as well.

So, anylocalid="1649438703970" $\mathrm{k}≤\mathrm{l}$will suffice. Assume thatlocalid="1649438710562" $\mathrm{k}<\mathrm{l}$.

In this scenario,localid="1649438688537" ${\mathrm{N}}_1=\mathrm{k}$,${\mathrm{N}}_2=\mathrm{l}$or${\mathrm{N}}_2=\mathrm{k}$,${\mathrm{N}}_1=\mathrm{l}$must be used.

As a result, there are only two options:

Hence

$\mathrm{P}(\mathrm{X}=\mathrm{k},\mathrm{Y}=\mathrm{l})=\frac{2}{36}$

If localid="1649438717904" $\mathrm{k}=\mathrm{l}$,localid="1649438724319" $({\mathrm{N}}_1,{\mathrm{N}}_2)=(\mathrm{k},\mathrm{k})$.

Hence,

$\mathrm{P}(\mathrm{X}=\mathrm{k},\mathrm{Y}=\mathrm{l})=\frac{1}{36}$