# Adding the probabilities of two events when one is a subset of the other

I'm trying to make some predictions about events happening in the next year, and I have a question about the methodology I'm using.

Here's a stylized example of the problem I'm working on:

Alice and Bob are running a marathon for the first time. I'm interested in estimating p(Alice or Bob finish the race) and p(Alice and Bob finish the race), as well as p(Alice finishes the race) and p(Bob finishes the race).

I've done some information gathering on Alice's and Bob's fitness, training schedule and the base rate of first-time entrants finishing marathons. After this research, I believe that p(Alice or Bob finish the race) = 0.75, and that p(Alice and Bob finish the race) = 0.35

To figure out p(Alice finishes the race) and p(Bob finishes the race), I'm considering adding up the two probability estimates I've already made, then divvying up the total probability between the two (based on the information I have about them). So this would look like: 0.75 + 0.35 = 1.10, divvy up as p(Alice) = 0.8 and p(Bob) = 0.3

However, I'm not sure if it is appropriate to add up p(Alice or Bob) and p(Alice and Bob), given that the second event is a subset of the first. Is this an appropriate method for making predictions?

Is there another way I could arrive at predictions for p(Alice) and p(Bob) using the information I have about p(Alice and Bob), p(Alice or Bob), and their backgrounds?

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Are the probabilities of Alice & Bob finishing the race independent of each other? If they are correlated, do you know p(A|B) or p(B|A)? – gung Jan 7 at 17:44
I'm assuming they're independent. – ioannes Jan 7 at 17:46

The inclusion-exclusion principle states $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Therefore, you know that $P(A\cup B)+P(A\cap B)=P(A)+P(B).$ Without further information/assumptions, it is not possible to uniquely identify $P(A)$ or $P(B).$

You use the word "conditional" in your title, but it's important to note that this is not a problem which contains conditional probabilities. A conditional probability is something of the form "What's the probability that Alice finishes given that Bob finishes?" The notation for this is $P(A|B),$ and the technology to work with that is Bayes rule, which is just one particularly prominent conditional probability relation.

Gung's comment points the way to a solution to identifying $P(A)$ and $P(B)$ using conditional probabilities. For example, if we know $P(B|A)$, we can use Bayes rule, $$P(B|A)=\frac{P(A\cap B)}{P(A)}$$ and we can solve for $P(A)$ using algebra.

You've commented that you're assuming independence. Independence is defined to mean that $P(A\cap B)=P(A)P(B).$ Since we know $P(A)+P(B)=P(A\cap B)+P(A\cup B)$ and also that $P(A\cap B)=P(A)P(B),$ the solution set is the set of points satisfying the following criteria:

1. $P(A)\in [0,1]$
2. $P(B)\in [0,1]$
3. $P(B)P(A)=P(A\cap B)$
4. $P(A)+P(B)=P(A\cap B)+P(A\cup B)$

An obvious way to solve this is to just graph the lines from (3) and (4) as functions of $P(A)$ and $P(B)$. The intersection is the answer.

One caveat about independence: Assuming independence is a very strong assumption. When the independence assumption is violated, it's usually the case that results are not "slightly wrong" but spectacularly wrong.

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Great, thanks for the clear answer. Using "conditional" in the title was sloppy language on my part – I'll endeavor to be more precise in the future :) – ioannes Jan 7 at 17:39
@ioannes you can edit the question title. It might be a good idea to help people search in the future. Also, if user777's answer is correct, you should mark it as such (click the checkmark on the left) – ssdecontrol Jan 7 at 18:00
Thanks – I'm a noob. Did those things. – ioannes Jan 7 at 18:02
@user777 Thanks for the edits; this is really helpful. – ioannes Jan 7 at 18:07