Question

Suppose that a die is rolled twice. What are the possible values and their corresponding probabilities that the following random variables can take on: (a) the maximum value to appear in both rolls; (b) the sum of the values obtained from the two rolls; (c) the value of the first roll minus the value of the second roll?

Answer 1

When dies is rolled twice The sample space are-

(a) The maximum value appears in both rolls is $(6,6)$ .

and probability $=$ $\dfrac{1}{36}$

(b) The sum of the values obtained from rolls are-

$2\rightarrow (1,1)\\3\rightarrow(2,1),(1,2)\\4\rightarrow (1,3),(2,2),(3,1)\\5\rightarrow (1,4),(2,3),(3,2),(4,1)\\6\rightarrow (1,5),(2,4),(3,3),(4,2),(5,1)\\7\rightarrow (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\\8\rightarrow (2,6),(3,5),(4,4),(5,3),(6,2)\\9\rightarrow(3,6),(4,5),(5,4),(6,3)\\10\rightarrow(4,6),(5,5),(6,4)\\11\rightarrow(5,6),(6,5)\\12\rightarrow(6,6)$

Probability that sum is $2=\dfrac{1}{36}$

Probability that sum is $3=\dfrac{2}{36}$

Probability that sum is $4=\dfrac{3}{36}$

Probability that sum is $5=\dfrac{4}{36}$

Probability that sum is $6=\dfrac{5}{36}$

Probability that sum is $7=\dfrac{6}{36}$

Probability that sum is $8=\dfrac{5}{36}$

Probability that sum is $9=\dfrac{4}{36}$

Probability that sum is $10=\dfrac{3}{36}$

Probability that sum is $11=\dfrac{2}{36}$

Probability that sum is $12=\dfrac{1}{36}$

(c) Value of the first roll minus the value of the second roll can hev the values 0,1,2,3,4,5

Probabilty that diffirence is $0=\dfrac{6}{36}$

Probabilty that diffirence is $1=\dfrac{10}{36}$

Probabilty that diffirence is $2=\dfrac{8}{36}$

Probabilty that diffirence is $3=\dfrac{6}{36}$

Probabilty that diffirence is $4=\dfrac{4}{36}$

Probabilty that diffirence is $5=\dfrac{2}{36}$