Roll two dice once. Let A be the event that the sum
of the two dice is 10...
a) P(A)=
Here are all 36 possible dice rolls with the
ones with sum 10 colored red:
(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)
Count the red ones, there are 3.
So P(A) is 3 out of 36 = 3/36 = 1/12
...and B be the event that at least
one of the dice is odd.
b) P(B)=
Here are all 36 possible dice rolls with the
ones that have at least one of the dice odd in red:
(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)
Count the red ones, there are 27.
So P(B) is 27 out of 36 = 27/36 = 3/4
c) Find the probability that the sum of the two dice is at least 9.
Here are all 36 possible dice rolls with the
ones with sum of the dice at least 9 colored red:
(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)
Count the red ones, there are 10.
So P(sum≥9) is 10 out of 36 = 10/36 = 5/18.
Edwin
