I am sitting here trying to set up a 2 x 2 table and keep coming up with funky results. The question is: Suppose that 30% of the population have characteristic A and 40% have charactersitic B. Assume that the probabilities are independent.

What is the probability that a person selected at random from this populateion will have both characteristics. I take this to be the union of
of A and B and believe that the answer is .3 + .4 = .7. This is clearly not correct. It is way to high.

What is the probability of having Characeristic A among people who have charteristice B?

What is the probability of haing Chaacteristic A among peope who do not have characteristic B?

The union of two events (call them A and B) means “Either A occurs, or B occurs, or both occur”.

The intersection of A and B means “A and B both occur”.

The probability of a union is not the same as an intersection. You need to figure out which of these two situations you’re looking for and then do the calculation. The math is obviously very simple in your example, however I’ll note that your calc for a union is wrong…suppose that P(A) = 0.8 and P(B) = 0.9, would P(AuB) = 1.7?

I’m not sure what you’re putting in your 2x2 box but if you know the probabilities of two independent events, the probability of both of them occurring is the product of the two probabilities. That should launch you in the right direction.

(Also note that it’s against board rules to ask for help with homework, in case that’s what this is.)

Actually, it is and it isn’t homework. I am following a couple of online (free) courses from MIT. So, I am out of school, and doing this for my own self interest. I have been doing well, but seem somewhat befuddled by this section of an Epidemiology course they are offering. Obviously, they provide no backup, given that they don’t even charge for the courses.

I’d look at the second question again (specifically, I’d do so keeping in mind your reasoning regarding the third question). In case this is the issue, note that the probability of A among people who have B isn’t the same as the probability of having both A and B (e.g., the probability of owning a computer is nearly 100% among people with an SDMB account, while the global probability of both owning a computer and having an SDMB account is tiny).

In general, this is true, but the OP said he was assuming that the probabilities were independent—which means that the probability of having A is the same regardless of whether you look at the general population or just at those who have B (or just those who don’t have B).

Yes, I agree. But look again at the answer he gave to question 2, and the mistaken reasoning I suspect he followed.

Let me revise my example to illustrate the issue I was talking about, even where dealing with independent events: Suppose we look at families with two children. The probability of both the older and the younger child being boys is 25%. However, the probability of the younger child being a boy is 50% among those whose older child is a boy. The global probability P(A & B) of A and B is not the same as the probability P(A | B) of A among those with trait B.

Let me make that last sentence a little clearer: The global probability P(A & B) of having both traits A and B is not the same as the probability P(A | B) of having trait A among those with trait B.

Say that you have 100 people all lined up into a 10 person by 10 person square. Horizontally, 30 of them will be A starting from the back. Vertically, 40 of them will be B, starting from the left.

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