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2 answers:
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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))
