algebra hw help

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 8

.

Event B

: The sum is not divisible by 3

 and not divisible by 5

.

Round your answers to two decimal places.

Answer 1

total possible outcomes: 6*6 = 36

For the first one you would do:

Possibilities for dice 1: 3,4,5,6

1 and 2 are not included because the max you can get is 6+2 = 8 and that is not greater than 8

Possibilities for dice 2: 3,4,5,6

Focus on first die iterations:

(3+6) = 9 , (4+5) = 9, (4+6) = 10, (5+5) = 10, (5+6)=11, (6+6) = 12 ---> 6

If you switch them so that you're first roll becomes your second roll you get another 6

so that becomes 12 possibilities

so the probability should be 12/36 = .333 = 33%

If you are counting without duplicates it would be 6/36

hope that makes sense! Happy Studying