In the rolling or two fair dice calculate the following

Question

P(Sum of the two dice is 9)=

P(sum of the two dice is 6)=

P(sum of the two dice is not 7)=

P(sum of the two dice is 5 or 8)=

P(sum of the two dice is not 4 and not 10)=

Answer 1

When rolling two dice, there are 36 possible outcomes:

1-1

1-2

1-3

1-4

1-5

1-6

2-1

2-2

.

.

.

6-4

6-5

6-6

You can achieve a sum of 9 in four ways:

3-6

4-5

5-4

6-3

So P(9) = 4/36 = 1/9

You can achieve a sum of 6 in five ways:

1-5

2-4

3-3

4-2

5-1

So P96) = 5/36

You can achieve a sum of 7 in six ways:

1-6

2-5

3-4

4-3

5-2

6-1

That means you would NOT get a 7 in 30 ways (36 - 6 = 30)

So the P(not 7) = 30/36 = 5/6

You can get either a 5 or 8 in nine ways:

1-4

2-3

3-2

4-1

2-6

3-5

4-4

5-3

6-2

So P(5 or 8) = 9/36 = 1/4

For not 4 AND not 10, it may be easier to consider this as not (4 or 10)

The sum of 4 is achieved from 1-3, 2-2, 3-1 so 3/36 or 1/12

P(10) is the same because 10 is achieved with 4-6, 5-5, 6-4 so also 1/12

P(4 or 10) is 1/12 + 1/12 = 2/12 = 1/6

So P(not(4 or 10)) is 5/6