Please help with?

Event A: Their sum is greater than 7

We can solve through complementary events.
P(sum greater than 7) = 1- P(sum equal to 7)

What are the numbers that equal to 7?

1,6
2,5
3,4
4,3
5,2
6,1

In each case, you have `1/36` chance of rolling those two numbers. Here's why:

Looking at 1,6

You have `1/6` chance of rolling a 1 and you also have `1/6` chance of rolling a 6. Since these two chances are independent events, then you multiply the numbers together. `1/6times1/6=1/36`

The `1/36` applies to all 6 numbers. So `1/36times6 = 1/6`

Therefore, P(sum equal to 7) = `1/6`

So, P(sum greater than 7) = `1- 1/6 = 5/6`

Event B: the sum is not divisible by 4 and not divisible by 6

The largest sum you can have between two numbers is 12 ie 6+6 and the smallest number you can have is 2 ie 1+1

So numbers divisible by 4 in that range are: 4, 8, 12
and numbers divisible by 6 in that range are: 6, 12

Therefore, looking at the table below, you can see that there are 14 combinations that you cannot throw. Since each combination ie 1,3 or 6,6 and you have `1/36` chance of throwing each combination, then P(sum is divisible by 4 and 6) is equal to`1/36times14 = 7/18`

P(sum not divisible by 4 and 6) = 1-P(sum is divisible by 4 and 6)
P(sum not divisible by 4 and 6) = `1-7/18 = 11/18`

![https://useruploads.socratic.org/Fg5eZ6JsQ8GC11vrgJ6g_sum_of_2_dice.jpg)

Consider all the possible results of two dice rolls.

`1+1=2`
`1+2=3`
`cdots`
`2+1=3`
`2+2=4`
`cdots`
`6+5=11`
`6+6=12`

In total, there are `36` combinations:

`2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12`

`15` out of these `36` are strictly greater than 7.
`22` out of these `36` are not divisible by 4 or by 6.

Therefore,

`P(A)=15/36=5/12`

`P(B)=22/36=11/18`