Chapter 6, Problem 6.1P

Two fair dice are rolled. Find the joint probability mass function of X and Y when

a. X is the largest value obtained on any die and

Y is the sum of the values;

b. X is the value on the first die and

Y is the larger of the two values;

c. X is the smallest and

Y is the largest value obtained on the dice.

To determine

To find: Joint probability mass function with X be the largest value Y be the sum of values.

Answer to Problem 6.1P

Joint probability mass function:

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

Explanation of Solution

Given information:

While rolling two fair dice,

X is the largest value obtained on any die.

Y is the sum of the values.

Let

N₁ and N₂ as the random variables that mark numbers obtained on the first and second die.

We know that

N₁ and N₂ are independent.

Such that

N₁ , N₂ ? D Unif(1,…,6).

In this part,

We have

  $X=\max \left(N_1,N_2\right)$

And

  $Y=N_1+N_2$

Thus,

  $X\in \left\{1,\mathrm{...},6\right\}$

And

  $Y\in \left\{2,\mathrm{...},12\right\}$

Also,

We have

  $X<Y$ almost certainly.

Then

Take any $k<l$ ,

Where,

k and l are from the ranges of X and Y .

Consider event

  $X=k,Y=l$ .

That means

The maximum value on any die is k .

And

The sum of both dice is l .

Now,

Observe that

If $l<2k$ ,

The only possible pairs of $\left(N_1,N_2\right)$ corresponding to the event are $\left(l,l-k\right)$ and $\left(l-k,k\right)$ .

If $l=2k$ ,

The only possible pair is $\left(k,k\right)$ .

Therefore,

The required probability mass function is

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

To determine

To find: Joint probability mass function with X be the value on first die and Y be the larger value.

Answer to Problem 6.1P

Joint probability mass function:

  $P\left(Y=l|X=k\right)P\left(X=k\right)=\{\begin{array}{c}\frac{k}{36}k=l\\ \frac{1}{36}k<l\end{array}$

Explanation of Solution

Given information:

While rolling two fair dice,

X is the value on the first die.

Y is the larger of the two values.

Let

N₁ and N₂ as the random variables that mark numbers obtained on the first and second die.

We know that

N₁ and N₂ are independent.

In this part,

We have

  $X=N_1$

And

  $Y=\max \left(N_1,N_2\right)$

Then

Observe that

  $\left\{1,\mathrm{...},6\right\}$

And

  $X\le Y$ almost certainly.

Take any $k\le l$ from the range $\left\{1,\mathrm{...},6\right\}$

Then

We have

  $P\left(X=k,Y=l\right)=P\left(Y=l|X=k\right)P\left(X=k\right)$

Suppose that

  $k=l$

We already have

  $X=k$

In such case,

N₂ can be any number from the range $1,\mathrm{...},k$ to obtain the required $Y=l$ .

Thus,

  $P\left(Y=l|X=k\right)P\left(X=k\right)=\frac{k}{6}\cdot \frac{1}{6}=\frac{k}{36}$

If $k<l$ ,

And

  $X=k$ ,

N₂ must be equal to l to obtain $Y=l$ .

Thus,

  $P\left(Y=l|X=k\right)P\left(X=k\right)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$

To determine

To find: Joint probability mass function with X be the smallest and Y be the largest value obtained.

Answer to Problem 6.1P

Joint probability mass function:

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

Explanation of Solution

Given information:

While rolling two fair dice,

X is the smallest value.

Y is the largest value.

Let

N₁ and N₂ as the random variables that mark numbers obtained on the first and second die.

We know that

N₁ and N₂ are independent.

In this part,

We have

  $X=\min \left(N_1,N_2\right)$

And

  $Y=\max \left(N_1,N_2\right)$

We also have

  $X\le Y$ almost certainly.

Then

Take any $k\le l$ .

Suppose that

  $k<l$

In such case,

We need to have

  $N_1=k,N_2=l$

Or

  $N_1=l,N_2=k$

Thus,

There are only two possibilities.

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

If $k=l$ ,

The only possibility will be $\left(N_1,N_2\right)=\left(k,k\right)$ .

Thus,

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