# Addition Rule and Conditional Probabilities

My question comes from part of an assignment question that I am having difficulty understanding.

I have a community of people, males and females split into three age categories (A, B, and C), so six groups altogether (MA, MB, MC, FA, FB, and FC). The probability of having a particular medical condition for each group has been given e.g. P(MA)=0.015.

I'm told that I have three individuals, selected at random from the community. One male from age group A (MA), one female from age group B (FB), and one female from age group C (FC).

I have to calculate two probabilities from the information I have. Firstly I need to calculate the probability that one of the three individuals selected has the medical condition. Now to calculate this, is it as simple as summing the probabilities that an individual from the groups MA, FB, & FC has the condition? This makes sense to me, but it seems too simple.

Secondly, I'm told that exactly one of the individuals selected has the condition. I'm asked to calculate the conditional probability that the affected individual is female.

For this, I defined A as being female and B as having the condition. Now if I'm thinking this through correctly, I am being asked what the probability of being female is, given an individual has the condition.

So if one of the three has the condition, the P(B) should be 1/3, and P(A AND B) should be the sum of P(FB) and P(FC)? (from the first part of the question). So I should be able to use P(AǀB) = P(A AND B)/P(A). I have done it this way but my answer feels too low (less than 0.25). Perhaps I am tired and I have been looking at it for too long now, but I am terribly confused...

I could give the actual values, but I'm more looking for help in how to think about the problem rather than have it calculated for me. I hope my description is clear enough and that someone can help straighten me out..

Thank-you!

D

Let $$X$$, $$Y$$ and $$Z$$ be the random variable that the respective individual has the condition. Then $$P(X=1)$$ is the probability that the individual $$X$$ has the condition and $$P(X=0)=1-P(X=1)$$ that he hasn't.

First Question: What is the probability that exactly one subject has the condition? This is given by $$P(X+Y+Z=1)=P(X=1,Y=0,Z=0)+P(X=0,Y=1,Z=0)+P(X=0,Y=0,Z=1)$$

Second question: What is the probability that female individual ($$Y \vee Z$$) has the condition given that exactly one of the three individuals has the condition. This is given by

$$P(Z \vee Y| X+Y+Z=1)= P((Z \vee Y) \wedge (X+Y+Z=1))/P(X+Y+Z=1) = (P(X=0,Y=1,Z=0)+ P(X=0,Y=0,Z=1))/ P(X+Y+Z=1)$$

Hints:

! Assume $$P(X,Y,Z)=P(X)P(Y)P(Z)$$

! $$P(X)=P(\mathrm{MA})$$