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 succes

Question

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 7

Event
B
: The sum is divisible by 4

Answer 1

15 ways to get 8, 9, 10, 11, or 12 over 36 possible outcomes = 15/36 = 5/12

1/4 for 2nd part

divisible by 4 means 1/4 of the possible outcomes

Answer 2

Hi Cynthia!

 

The sample space of rolling two dice has 36 possible outcomes.  Remember that the "sample space" is a set which contains all possible outcomes.   

 

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

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

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

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

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

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

 

The probability of rolling a 2 is 1/36.

The probability of rolling a 3 is 2/36.

The probability of rolling a 4 is 3/36.

The probability of rolling a 5 is 4/36.

The probability of rolling a 6 is 5/36. 

The probability of rolling a 7 is 6/36.

The probability of rolling an 8 is 5/36.

The probability of rolling a 9 is 4/36.

The probability of rolling a 10 is 3/36.

The probability of rolling an 11 is 2/36.

The probability of rolling a 12 is 1/36.

 

Using these probabilities, you can answer your questions:

 

Probability of rolling a number bigger than 7 is the sum of the probabilities of rolling an 8, 9, 10, 11, and 12.

5/36 + 4/36 + 3/36 + 2/36 + 1/36 = 15/36 which reduces to 5/12

 

Probability of rolling a number divisible by 4 is the sum of the probabilities of rolling a 4, 8, and 12.

3/36 + 5/36 + 1/36 = 9/36 which reduces to 1/4.

 

Hope this helps.