Question

How do I do this please?

Answer 1

`P(alpha) =5/12`, `P(beta)=11/18`

Explanation:

The possible sums are: `2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12`

Therefore the total number of possible sums is `11`. However, the number of ways to arrive at particular total differs.

E.g. To reach a total of 2 is only possible 1 way - 1 and 1
but a total of 6 can be reached in 5 ways - 1 and 5, 5 and 1, 2 and 4, 4 and 2, 3 and 3.

Mapping out all the possible ways to reach a given sum yields the following.

Sum `->` No of Ways
2 `->` 1
3 `->` 2
4 `->` 3
5 `->` 4
6 `->` 5
7 `->` 6
8 `->` 5
9 `->` 4
10 `->` 3
11 `->` 2
12 `->` 1

So, the total number of ways any outcome can be achieved is:

`(1+2+3+4+5)xx2 +6 =36`

Since the dice is"fair" then each outcome is equally likely. Therefore to find the probability of an event we can take the number of sums that satisfy the event criteria, compute the number of ways that these can be achieved and divide by 36.

Event `alpha` - Sum is greater than 7

Sums that satisfy the event criteria are: `8-12` inclusive.

Number of ways to achieve these: `5+4+3+2+1 = 15`

`-> P(alpha) = 15/36 = 5/12`

Event `beta` - Sum is not divisible by 4 and not divisible by 6
(Assume the result must be an integer)

Sums that satisfy the event criteria are: `2, 3, 5, 7, 9, 10, 11` .

Number of ways to achieve these: `1+2+4+6+4+3+2 = 22`

`-> P(beta) = 22/36 = 11/18`