Question

Let a random experiment be the casting of a pair of fair dice, each having six faces, and let the random variable X denote the sum of the dice. (a) With reasonable assumptions, determine the pmf f (x) of X. Hint: Picture the sample space consisting of the 36 points (result on first die, result on second die), and assume that each has probabilit Find the probability of each possible outcome of X, namely, x = 2, 3, 4, . . . , 12. b) Draw a probability histogram for f(x). y 1/36.

Answer 1

Answer:

a)

Pr(2) = 1/36

Pr(3) = 2/36 = 1/18

Pr(4) = 3/36 = 1/12

Pr(5) = 4/36 = 1/9

Pr(6) = 5/36

Pr(7) = 6/36 = 1/6

Pr(8) = 5/36

Pr(9) = 4/36 = 1/9

Pr(10) = 3/36 = 1/12

Pr(11) = 2/36 = 1/18

Pr(12) = 1/36.

b) See attached

Step-by-step explanation:

Since the dice are fair: We expect the following possible outcomes:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

And sum gives us the following:

2,3,4,5,6,7

3,4,5,6,7,8

4,5,6,7,8,9

5,6,7,8,9,10

6,7,8,9,10,11

7,8,9,10,11,12

Therefore, the probability of each sum outcome is given as follow:

Pr(x = sum outcome of event) = number of event/total number of possible events. Thus:

Pr(2) = 1/36

Pr(3) = 2/36 = 1/18

Pr(4) = 3/36 = 1/12

Pr(5) = 4/36 = 1/9

Pr(6) = 5/36

Pr(7) = 6/36 = 1/6

Pr(8) = 5/36

Pr(9) = 4/36 = 1/9

Pr(10) = 3/36 = 1/12

Pr(11) = 2/36 = 1/18

Pr(12) = 1/36.

For the attached plot, see the R code below:

probability = c(1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36)

hist(probability, prob=T)

lines(density(probability))