Answer:
P(A) = 13/18
P(B) = 7/18
Step-by-step explanation:
A) Sum greater than 5
If the sum needs to be >5, then it can be 6, 7, 8, 9, 10, 11, and 12.
There are 26 out of 36 events where the sum is >5, thus P(A) = 26/36 = 13/18.
B) Sum divisible by 4 or 6
The sum ranges from 2 to 12, there are two methods to calculate this:
B1) Method 1: Calculate the frequency one by one
The sums divisible by 4 or 6 are S = {4, 6, 8, 12}. There are 9 of them, n(S) = 14. Thus P(B) = 14/36 = 7/18.
B2) Method 2: Use the rule of not mutually exclusive events
Not mutually exclusive events (might also be called mutually inclusive) happens when two events can happen together at the same time. In this problem, the sum might be divisible by 4 and 6 at the same time, i.e. the number 12.
The common formula for not mutually exclusive events is P(A∪B) = P(A) + P(B) - P(A∩B). Nb: P(A) isn't talking about the previous problem, nor does P(B) talk about the second section as a whole; the terms P(A) and P(B) will be utilized next, explained with their uses.
P(A) = probability of the sum being divisible by 4, i.e. the number 4, 8 and 12
P(A) = (3+5+1) / 36 = 9/36 = 1/4
P(B) = probability of the sum being divisible by 6, i.e. the number 6 and 12
P(B) = (5+1) / 36 = 6/36 = 1/6
P(A∩B) = probability of the sum being divisible both by 4 and 6, i.e. the number 12
P(A) = 1 / 36 = 1/36
Here comes the use of formula P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = 1/4 + 1/6 - 1/36 = 7/18
Nb: As you can see, both method B1 and B2 give the same answer. Method B1 is usually easier to use when facing simpler problems like this. I'd recommend using method B2 for more complex and difficult questions, such as when you can't visualize the problem with a table because there are too much sets of variables.
Image of table for reference: https://qph.fs.quoracdn.net/main-qimg-1283a98bd649f522986e2f91a592ffd3
