An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (repr

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 2 and not divisible by 6 . Write your answers as exact fractions.

(a)

(b)

Explanation / Answer

for the first roll of the die you can get 1, 2, 3, 4, 5, or 6

for the second roll of the die you can get 1, 2, 3, 4, 5, or 6

Since there are six numbers on each die, there are 6 X 6 = 36 total different results

when you roll two dice which is the same thing as rolling one die twice in succession.

There is exactly 1 way to get a total of 2: 1 and 1
There are two ways to get 3: 1 and 2, 2 and 1
There are three ways to get 4: 1 and 3, 3 and 1, 2 and 2
four ways to get 5
five ways to get 6
six ways to get 7
five ways to get 8
four ways to get 9

three ways to get 10
two ways to get 11
one way to get 12

EVENT A:

for the sum to be greater than 8,

it must be, 9, 10, 11, or 12 and there are 4 plus 3 plus 2 plus 1 ways to get those numbers

out of the total of 36.

=> (4+3+2+1)/36

=10/36 ===> 5/18

Event B:

The only multiples of 6 possible are 6 and 12. which are divisible by 2 and divisible by 6,

So their probability is,

==> (1+5)/36= 6/36 =1/6

And the question asked what is the probability of not getting a multiple of six so we subtract the above probability from 1 which equals 5/6

Probability of sum not divisible by 2 and not divisible by 6 is,

==> 1-1/6===> 5/6

Answer.

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