# Simple probability question

A person from group A has 20% chance of having some characteristic

A person from group B has 30% chance of having the same characteristic

How can I calculate the probability of a person belonging to both groups having the given characteristic?

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Since a number of people pointed out that it's impossible to know, I'll change my question.

Let's assume that there are 10 different groups. Do I need to know the probabilities for each possible combination, or can I infere at least some probabilities?

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I've added a solution witch I think is plausible for some cases.

• Given this information, you can't do this. Commented Sep 27, 2012 at 20:31
• What additional information do I need? If you want some context, think disease risk assesment Commented Sep 27, 2012 at 20:32
• I think the only that will help is observations of people from both groups. There's no theoretical way of calculating the interaction effect - the answer could be anything between 30% and 100%. Commented Sep 27, 2012 at 20:35
• Essentially, you'd need to know what you are asking. From what you have, there is no way to know if the answer should be something between 20 and 30, or something greater than 30, or even, possibly something less than 20. Risks can be additive, but they needn't be: There can be (and often is) an interaction. E.g. Radon exposure is related to lung cancer. Smoking is too. But the effects of radon are greater for smokers than nonsmokers. But it can go the other way, too. Commented Sep 27, 2012 at 20:36
• @Glen_b no, that's a mistake from me, the interaction could be anything down to zero. Commented Sep 28, 2012 at 11:59

PeterFlom is right that the information you have is not enough to answer the question. However, when you ask "what do I need to know", I could say that if a divine spirit told you $$P(A\cup B)=.6$$ then since $$P(A)=.2$$ and $$P(B)=.3$$, then the desired answer .1. It comes from the well-known formula in probability.

For any two event A and B, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$. So $$P(A\cap B)=P(A)+P(B)-P(A\cup B)$$

So you only need to know $$P(A\cup B)$$. This also makes it clear why you can't solve with the information at hand You do need to know $$P(A\cup B)$$

I am adding to my answer because the OP mentioned that his problem is considering information where A=[set of radiologists in the sample space] and B=[set of members of the sample space with cancer] and he wants to understand what is the probability that a member of the cancer set that also is a radiologist will have cancer. He states that he thinks that the probability that a cancer patient who is a radiologist would have a probability of 0.1 for having cancer. He thinks that is too low and presumably even 0.2 might seem too low as well. I have 2 responses to that.

1. It seems that the answer to that question is asking for $$P(B|A)$$

$$P(B|A)=P(A\cap B)/P(A)$$. This does not have the same bounds as $$P(A\cup B)$$.

For this problem $$P(B|A)$$ is not known either since we do not know $$P(A\cap B)$$. In this case $$P(B|A)=P(A\cap B)/0.2 =5 P(A\cap B)$$.

So $$P(B|A)$$ has a lower bound of 0 and an upper bound of 5(.2)=1! So $$P(B|A)$$ ** can be any probability regardless of what $$P(A)$$ and $$P(B)$$ are!**

2. Why is a value less than 0.2 plausible?

This is just a theory, but given no information about the smoker's profession, the probability he/she has cancer is 0.2. Now radiologists understand the dangers of cancer better than most smokers who are not radiologists. So the radiologists who smoke might tend to be light smokers. Now light smokers are less likely to have cancer because smokers with cancer predominantly have lung cancer and light smokers are less likely to have lung cancer than moderate or heavy smokers.

Initially, it was not clear to me that the OP wanted $$P(B|A)$$. After explaining the problem I think that it is because he is asking about the probability of cancer when you know the individual is a radiologist.

• I think what I need is something else. It makes little sense for the answer to be less than .3 in my context. I'm looking for models suitable for risk assessment, and an example maximally similar to mine. I'm out of my water to be more specific. Commented Sep 27, 2012 at 22:35
• This result is a correct formula for any two events. What models are you talking about. There seems to be details about this problem that you haven't told us about. With the information you gave us P(AUB) can have a range of values. I said that you can't know P(AUB) and 0.6 was just a hypothetical value. If you know P(A)=0.2 and P(B)=0.3 then 0.0<=P(A∩B)<=0.2 and 0.3<=P(AUB)<=0.5. If the answer can't be in the interval [0. 0.2] then either P(A) is not equal to 0.2 or P(B) is not 0.3 or they both are different from what you specified. Commented Sep 28, 2012 at 2:33
• There is no other possibility without violating the laws of probability! Commented Sep 28, 2012 at 2:33
• That does not seem weird to me. That makes perfect sense to me. The smoking radiologist form a subset of both the set of radiologists and the smokers. So P(A∩B) has to be no bigger than the minimum of P(A) and P(B). Commented Sep 28, 2012 at 8:58
• -1 for deducing from $P(A) = 0.2$, $P(B) = 0.3$, and the claim of a divine spirit that $P(A\cup B) =0.6$ that $P(A\cap B)=0.1$ instead of questioning said allegedly heavenly claim, and for not bothering to correct the error even when it has been pointed out. Commented Dec 27, 2012 at 23:14

Suppose a person in a car accident has 30% chance of dying.
Suppose a person who drank detergent has 20% chance of dying.
Meaning it's 70% and 80% survival probabilities respectively.
There is 70% * 80% = 56% chance that a person who did both, will survive twice.
Meaning there is 44% chance he will die.

It does get silly if there are many groups with a percentage above 0, or if there is interaction, but it seems to work here.

• Drinking detergent may affect your driving. Commented Oct 28, 2012 at 23:10
• or, someone who has car accidents and also drinks detergent may be subject to an unmeasured but significant third characteristic that leads to other unusual behaviour. Commented Jan 27, 2013 at 8:15