# Two Events A & B: Is r+s >1 Possible? - Brendan's Query

• brendan
In summary, the probability of two events A and B happening together, p(A + B), is always less than or equal to the sum of their individual probabilities, p(A) + p(B), regardless of whether they are mutually exclusive or not. Additionally, if A and B are not mutually exclusive, p(A + B) can be greater than 1, but this does not contradict the fact that p(A) + p(B) cannot be greater than 1.
brendan
I have read a question which has

Two events A and B such that p(A) = r and p(B) = s with r,s >0 and
r + s > 1

My querry is about r + s > 1 is this possible ? I thought that it had to be < or = to 1

regards
Brendan

Suppose you have a die, and you throw it once. Let A be "the outcome is 1, 2, 3, 4 or 5" and B "the outcome is 3, 4, 5 or 6".

Note however, that though p(A) + p(B) > 1, you always have p(A + B) <= 1 (with p(A + B) being the probability of A or B happening).

brendan said:
I have read a question which has

Two events A and B such that p(A) = r and p(B) = s with r,s >0 and
r + s > 1

My querry is about r + s > 1 is this possible ? I thought that it had to be < or = to 1

regards
Brendan

If A and B are "mutually exclusive" events, then P(A+ B)= P(A)+ P(B). Since P(A+B) cannot be larger than 1, neither can P(A)+ P(B).

But if A and B are NOT "mutually exclusive" that is not true. Suppose you roll a single die. Let A be "the number on the die is larger than 2" and B is "the number on the die is even"
Since there are 4 numbers on the die larger than 2, P(A)= 4/5= 2/3. Since there are 3 even numbers on the die, P(B)= 3/6= 1/2. P(A)+ P(B)= 2/3+ 1/2= 7/6> 1.

Of course, that is NOT P(A+B). The numbers on the die that are "either larger than 2 or even" are 2, 3, 4, 5, 6 so P(A+ B)= 5/6.

"Larger than 3 or even" are not mutually exclusive. 4 and 6 are both "larger than 3" and even. P(A and B)= 2/6= 1/3 so P(A+ B)= P(A)+ P(B)- P(A and B)= 2/3+ 1/2- 1/3= 7/6- 2/6= 5/6.

Thanks very much guys.
I see that it makes sense that they could both equal > 1 if both are not mutually exclusive.

Now I just have to prove it !

CompuChip said:
Note however, that though p(A) + p(B) > 1, you always have p(A + B) <= 1 (with p(A + B) being the probability of A or B happening).

Don't you mean:

p(A + B) <= P(a)+P(b)

which is a triangle inequality.

Yes, that is also true. Actually I think it is something like
$$p(A \cup B) = p(A) + p(B) - p(A \cap B)$$
which shows both the triangle inequality and the statement about mutually exclusive events (where $A \cap B$ is empty).

However, the point I wanted to stress is that the probability of A or B is always less than or equal to one, as you'd expect, contrary to the probability of A plus the probability of B; therefore p(A) + p(B) is not the same as p(A + B).

## What are the two events A and B in Brendan's query?

The events A and B refer to two separate events that are being studied in relation to each other. In this case, the events A and B could represent the occurrence of two different variables or phenomena in a scientific experiment or study.

## What does "r+s" represent in the equation?

"r+s" represents the combined probability or occurrence of the two events A and B happening together. This is known as the joint probability and is a common concept in statistics and probability theory.

## Is it possible for r+s to be greater than 1?

No, it is not possible for r+s to be greater than 1. The sum of probabilities for all possible outcomes must equal 1, as it represents the total probability of all possible events occurring. If r+s were to be greater than 1, it would violate this fundamental principle of probability.

## What does a value of r+s less than 1 indicate?

A value of r+s less than 1 indicates that the two events A and B are not mutually exclusive, meaning they can occur together. This means that the probability of both events happening at the same time is less than the probability of either event happening alone.

## How is the joint probability calculated in this scenario?

The joint probability of events A and B is calculated by multiplying the individual probabilities of each event. In other words, the joint probability of A and B occurring together is equal to the probability of A multiplied by the probability of B.

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