Tickbox Quiz #15 (Independent vs Mutually Exclusive events)

[Archi]

I am getting confused in these two concepts when we talk about if their sum is equal to 1 or not. In the question below, Greg says that if sum of p+q is 1.3 they could be independent but how is that possible because shouldn’t the sum of their probabilities equal to 1.

Then another contrast is here-

Here while assuming numbers for p and q, Greg chose numbers in a way so that sum of independent probabilities come under 1. Here he puts independent case equals to 1 and then finds out the value of both given 0.7 and 0.5.

So, my question is when do I know that sum of independent probabilities can be greater than 1 or less than 1 and for mutually exclusive are is the sum of the probabilities always going to be 1?

[Leaderboard]

@Archi

Greg says that if sum of p+q is 1.3 they could be independent but how is that possible because shouldn’t the sum of their probabilities equal to 1

No - recall the inclusion-exclusion formula. That could also answer your second point as well.

[Archi]

okay let me know if I understood your point correctly-

1. In the first picture, the sum of p + q is 1.3 which means they could be independent because overlap can exist and we are saying this with certainty because sum of p + q is greater than 1 so when we will subtract PnQ case from it it will be 1 or less than 1 which wouldn’t be possible if these were mutually exclusive events as we won’t have anything to subtract from them. So, logically this has to be an independent case for it to be under 1 or less than 1. Is this inference correct ?

2. But what I am still unable to understand is how can we put 0.7 + 0.5 -both equal to 1 (the equation that Greg wrote in the screenshot) because that would mean probability of AUB should always be equal to one considering all cases P(A) + P(B) -P(AnB) + neither which logically seems correct but then if we look at the question below and its solution 0.4 is incorrect and the way I calculated this is-

P(AUB) = P(A) + P(B) -P(AnB)
1 = 0.8 + P(B) - 0.2
that gives P(B) 0.4

Why is the above solution incorrect because in the solution video Greg solved it by-
P(A)P(B) = 0.2
0.8P(B) = 0.2
P(B) = 0.25

So, my question is why is our logic for these two calculations not same like for Dakota’s case when we put AUB = 1 to find P(AnB) ?

[Leaderboard]

@Archi

• In the first picture, the sum of p + q is 1.3 which means they could be independent because overlap can exist and we are saying this with certainty because sum of p + q is greater than 1 so when we will subtract PnQ case from it it will be 1 or less than 1 which wouldn’t be possible if these were mutually exclusive events as we won’t have anything to subtract from them. So, logically this has to be an independent case for it to be under 1 or less than 1. Is this inference correct ?

It could, but doesn’t have to be.

@Archi

1. But what I am still unable to understand is how can we put 0.7 + 0.5 -both equal to 1 (the equation that Greg wrote in the screenshot) because that would mean probability of AUB should always be equal to one considering all cases P(A) + P(B) -P(AnB) + neither which logically seems correct but then if we look at the question below and its solution 0.4 is incorrect and the way I calculated this is-

This is because we are specifically trying to minimise P(A \cap B) in this case.

@Archi

P(AUB) = P(A) + P(B) -P(AnB)
1 = 0.8 + P(B) - 0.2
that gives P(B) 0.4

You do not know what P(A \cup B) is other than the fact that it is less than or equal to 1.