Question

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

Short answer

1. The probability of maximum value to appear in the two rolls is $P=\frac{11}{36}$
2. The probability of minimum value to appear in the two rolls is $P=\frac{1}{36}$
3. The probability of the sum of the two rolls is $P=\frac{1}{36}$
4. The probability of the sum of the value of the first roll minus the value of the second roll is$P=\frac{1}{36}$

Step-by-step solution

Given Information (Part-a)

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.

Calculate the Probability of the Maximum Value to appear in the Two rolls (Part-a)

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}$

Final Answer (Part-a)

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

Given Information (Part-b)

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 (Part-b)

Substitute the values of the first roll into $X$

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

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

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

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

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

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

Probability Table (Part b)

$X$	1	2	3	4	5	6
$P\left(X\right)$	$\frac{11}{36}$	$\frac{9}{36}$	$\frac{7}{36}$	$\frac{5}{36}$	$\frac{3}{36}$	$\frac{1}{36}$

Final Answer (Part-b)

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

Given Information (Part c)

Let's consider that a die is rolled twice.

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

Calculate the Probability of the sum of the two rolls  (Part c)

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 table (Part c)

$X$	2	3	4	5	6	7	8	9	10	11	12
$P\left(X\right)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

Final Answer (Part c)

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

Given Information (Part d)

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

Calculate the sum of the value of the first roll minus the value of the second roll (Part d)

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

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

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

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

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

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

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

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

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

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

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

Final Answer (Part d)

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