Q. 6.1 - A First Course in Probability

Question

Two fair dice are rolled. Find the joint probability mass function of $\mathrm{X}$and $\mathrm{Y}$ when

(a) $\mathrm{X}$is the largest value obtained on any die and$\mathrm{Y}$ is the sum of the values;

(b) $\mathrm{X}$ is the value on the first die and $\mathrm{Y}$ is the larger of the two values;

(c) $\mathrm{X}$ is the smallest and $\mathrm{Y}$ is the largest value obtained on the dice.

Step-by-Step Solution

Answer

a. Joint probability mass function 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. Joint probability mass function is $\frac{\mathrm{k}}{36}$,$\frac{1}{36}$.

c. Joint probability mass function is $\frac{2}{36}$,$\frac{1}{36}$.

Step 1: Calculation for join probability mass function (part a)

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 that${\mathrm{N}}_1,{\mathrm{N}}_2~\mathrm{DUnif}(1,.....,6)$.

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

As a result,$\mathrm{X}\in \{1,.......,6\}$ and$\mathrm{Y}\in \{2,........,12\}$.

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

Take any $\mathrm{k}<\mathrm{l}$,where $\mathrm{k}$and$\mathrm{l}$are from the above-mentioned ranges.

Consider the $\mathrm{X}=\mathrm{k}$,$\mathrm{Y}=\mathrm{l}$occurrence.

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

Note that the only conceivable pairs of $({\mathrm{N}}_1,{\mathrm{N}}_2)$are $(\mathrm{k},\mathrm{l}-\mathrm{k})$and$(\mathrm{l}-\mathrm{k},\mathrm{k})$

for $\mathrm{l}<2\mathrm{k}$

$(\mathrm{k},\mathrm{k})$is possible for$\mathrm{l}=2\mathrm{k}$

As a result, the needed PMF is

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

Step 2: Joint probability mass function (part b)

b.

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

Both variables are in the$\{1,......,6\}$ range.

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

Assume that$\mathrm{k}=\mathrm{l}$and that $\mathrm{X}=\mathrm{k}$is handed to us. In that situation,${\mathrm{N}}_2$can be any number between$1,.......,\mathrm{k}$to get the requisite$\mathrm{Y}=\mathrm{l}$.

Hence

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

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

If$\mathrm{k}<\mathrm{l}$and we are given that $\mathrm{X}=\mathrm{k},{\mathrm{N}}_2$must be equal to$\mathrm{l}$to obtain$\mathrm{Y}=\mathrm{l}$.

So, in that case

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

$=\frac{1}{36}$

Step 3: Calculation for joint probability mass function (part c)

c.

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

We almost likely have$\mathrm{X}\le \mathrm{Y}$as well.

So, any$\mathrm{k}\le \mathrm{l}$will suffice. Assume that$\mathrm{k}<\mathrm{l}$.

In this scenario,${\mathrm{N}}_1=\mathrm{k}$,${\mathrm{N}}_2=\mathrm{l}$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 $\mathrm{k}=\mathrm{l}$,$({\mathrm{N}}_1,{\mathrm{N}}_2)=(\mathrm{k},\mathrm{k})$.

Hence,

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