Question 1084318: An ordinary (fair) die is a cube with the numbers
1
through
6
on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
Compute the probability of each of the following events:
Event
A
: The sum is greater than
9
.
Event
B
: The sum is divisible by
2
.
Write your answers as exact fractions.
Found 2 solutions by rothauserc, Boreal:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! There are 6 * 6 = 36 possible events
:
(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)
:
A. Sum is > 9
(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)
There are 6 events whose sum is > 9
Probability(P) sum is > 9 = 6/36 = 1/6
:
B. sum is divisible by 2
There are 36/2 = 18 events whose sum is divisible by 2
P sum is divisible by 2 is 18/36 = 1/2
:
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! It can be shown that the sum is greater than 9 (10, 11, 12) is 6/36 or 1/6. Those 6 possibilities are 4-6,5-5,6-4 and 5-6, 6-5, and 6-6
The sum is divisible by 2:
2: 1-1, or 1 way
4: 1-3, 2-2-, 3-1: 3 ways
6: 1-5, 2-4,3-3,4,2,5-1: 5 ways
8:also 5 ways 2-6, 6-2,3-5,5-3, 4-4
10: 3 ways
12: 1 way
That is a total of 18 ways out of 36 or 1/2.
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