Question

Caden rolls two fair number cubes numbered from 1 to 6. He first defines the sample space, as shown below:

(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)

Based on the sample space, what is the probability of getting a total of 6? (5 points)

a. 5 over 36
Selected:b. 6 over 36This answer is incorrect.
c. 7 over 36
d. 8 over 36

Answer 1

B. 6 over 36

Explanation:

Answer 2

B. 6 over 36

Explanation:

As we are given the sample space, the events that sum to 6 are:

{(1,5), (2,4), (3,3),(4,2),(5,1)}

Let n(S) be the number of items in sample space

n(S) = 36

Let E be the event the sum is 6

Then,

n(E) = 6

So,

P(Sum of two cubes is 6) = n(E) / n(S)

= 6/36

So, the correct answer is:

B: 6/36