If the die in Problem 4.7 is assumed fair, calculate the probabilities associated with the random variables in parts (a) through (d).

Given in the question that a die is assumed fair. We have to find the probabilities of the maximum value to appear in the two rolls.

The maximum value to appear in the two rolls

$\mathrm{X}=1(1,1)\mathrm{p}=\frac{1}{36}$

$X=2(2,1)(1,2)(2,2)$

$\mathrm{X}=3(1,3)(3,1)(2,3)(3,2)(3,3)\mathrm{p}=\frac{5}{36}$

$X=4(4,1)(1,4)(2,4)(3,4)(4,2)(4,3)(4,4)$$p=\frac{7}{36}$

role="math" $X=5(1,5)(5,1)(2,5)(5,2)(3,5)(5,3)(4,5)(5,4)$$(5,5)\mathrm{p}=\frac{9}{36}$

$X=6(1,6)(6,1)(2,6)(6,2)(3,6)(6,3)(4,6)(6,4)$$(5,6)(6,5)(6,6)p=\frac{11}{36}$

The probability of maximum value to appear in the two rolls is$P=\frac{11}{36}$

Given in the question that a die is assumed fair. We have to find the minimum value to appear in the two rolls;

Substitute the values of the first roll into $X$

$P(X=1)=P\left\{\right(6,1\left)\right(5,1\left)\right(4,1\left)\right(3,1\left)\right(2,1\left)\right(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,5\left)\right(1,6\left)\right\}=\frac{11}{36}$

$P(X=2)=P\left\{\right(6,2\left)\right(5,2\left)\right(4,2\left)\right(3,2\left)\right(2,2\left)\right(2,3\left)\right(2,4\left)\right(2,5\left)\right(2,6\left)\right\}=\frac{9}{36}$

$P(X=3)=P\left\{\right((6,3)(5,3)(4,3)(3,3)(3,4)(3,5)(3,6)\left)\right\}=\frac{7}{36}$

$P(X=4)=P\left\{\right((6,4)(5,4)(4,4)(4,5)(4,6)\left)\right\}=\frac{5}{36}$

$P(X=5)=P\left\{\right(6,5\left)\right(5,5\left)\right(5,6\left)\right\}=\frac{3}{36}$

$P(X=6)=P\left\{\right(6,6\left)\right\}=\frac{1}{36}$

The probability of minimum value to appear in the two rolls is$P=\frac{1}{36}$

Let's consider that a die is rolled twice.

We have to find the probability of the sum of the two rolls.

Let's compute the sum of the two rolls:

$P(X=2)=P\left\{\right(1,1\left)\right\}=\frac{1}{36}$

$P(X=3)=P\left\{\right(1,2\left)\right(2,1\left)\right\}=\frac{2}{36}$

$P(X=4)=P\left\{\right(1,3\left)\right(2,2\left)\right(3,1\left)\right\}=\frac{3}{36}$

$P(X=5)=P\left\{\right(1,4\left)\right(2,3\left)\right(3,2\left)\right(3,1\left)\right\}=\frac{4}{36}$

$P(X=6)=P\left\{\right(1,5\left)\right(2,4\left)\right(3,3\left)\right(4,2\left)\right(5,1\left)\right\}=\frac{5}{36}$

$P(X=7)=P\left\{\right(1,6\left)\right(2,5\left)\right(3,4\left)\right(4,3\left)\right(5,2\left)\right(6,1\left)\right\}=\frac{6}{36}$

$P(X=8)=P\left\{\right(2,6\left)\right(3,5\left)\right(4,4\left)\right(5,3\left)\right(6,2\left)\right\}=\frac{5}{36}$

$P(X=9)=P\left\{\right(3,6\left)\right(4,5\left)\right(5,4\left)\right(6,3\left)\right\}=\frac{4}{36}$

$P(X=10)=P\left\{\right(4,6\left)\right(5,5\left)\right(6,4\left)\right\}=\frac{3}{36}$

$P(X=11)=P\left\{\right(5,6\left)\right(6,5\left)\right\}=\frac{2}{36}$

$P(X=12)=P\left\{\right(6,6\left)\right\}=\frac{1}{36}$

The probability of the sum of the two rolls is$P=\frac{1}{36}$

Let's consider that a die is rolled twice.

We have to find the probability of the sum of the value of the first roll minus the value of the second roll

$P(X=5)=P\left\{\right(6,1\left)\right\}=\frac{1}{36}$

$P(X=4)=P\left\{\right(6,2\left)\right(5,11\left)\right\}=\frac{2}{36}=\frac{1}{18}$

$P(X=3)=P\left\{\right(6,3\left)\right(5,2\left)\right(4,1\left)\right\}=\frac{3}{36}=\frac{1}{12}$

$P(X=2)=P\left\{\right(6,4\left)\right(5,3\left)\right(4,2\left)\right(3,1\left)\right\}=\frac{4}{36}=\frac{1}{9}$

$P(X=1)=P\left\{\right(6,5\left)\right(5,4\left)\right(4,3\left)\right(3,2\left)\right(2,1\left)\right\}=\frac{5}{36}$

$P(X=0)=P\left\{\right(6,6\left)\right(5,5\left)\right(4,4\left)\right(3,3\left)\right(2,2\left)\right(1,1\left)\right\}=\frac{6}{36}=\frac{1}{6}$

$P(X=-1)=P\left\{\right(5,6\left)\right(4,5\left)\right(3,4\left)\right(2,3\left)\right(1,2\left)\right\}=\frac{5}{36}$

$P(X=-2)=P\left\{\right(4,6\left)\right(3,5\left)\right(2,4\left)\right(1,3\left)\right\}=\frac{4}{36}=\frac{1}{9}$

$P(X=-3)=P\left\{\right(3,6\left)\right(2,5\left)\right(1,4\left)\right\}=\frac{3}{36}=\frac{1}{12}$

$P(X=-4)=P\left\{\right(2,6\left)\right(1,5\left)\right\}=\frac{2}{36}=\frac{1}{18}$

$P(X=-5)=P\left\{\right(1,6\left)\right\}=\frac{1}{36}$

The probability of the sum of the value of the first roll minus the value of the second roll is$P=\frac{1}{36}$