# Conditional distribution and covariance problem

Let me start off by saying that I AM NOT ASKING YOU TO DO MY HOMEWORK FOR ME. Too often when I post questions like this people don't answer the question and tell me to do my own homework. I am not asking you to do it for me, I am asking you to help show me how it is done because I don't understand. So here is the problem I am struggling with:

Three patients are enrolled in a clinical trial. Each patient either has a complete response to initial treatment or not. The probability of a complete response for any patient is 0.6 and the response of each patient is independent of the others. Let X denote the number of complete responders out of the first two patients enrolled. Let Y denote the number of complete responders out of the second and third patients enrolled.

a) Find the conditional distribution of Y given X=1.

b) Find cov(X,Y).

I know how to find a conditional distribution and covariance in a general case, but I don't know what to do in this case. How do I set this problem up? Is the joint distribution multinomial, like so:

$$\frac{3!}{x!y!} (0.6)^x(0.6)^y$$

Or is it something else? I honestly don't know. Anyone give me a pointer?

There are just $2^3 = 8$ possible outcomes here, where an outcome tells whether or not each of the three patients is a complete responder. You can list them all, compute their probabilities and the values of $X$ and $Y$, and calculate everything using the definitions.

• I am having some trouble with this. First of all, how can there be 8 possible outcomes? So far as I can tell, there are only 7 possible outcomes. (X=0,Y=0), (X=1,Y=0), (X=0,Y=1),(X=1,Y=1),(X=2,Y=1),(X=1,Y=2), and (X=2,Y=2). What is the eighth outcome? – Abe Nov 3 '13 at 19:18
• Ah, I see. There are two different ways for (X=1,Y=1) to happen. – Abe Nov 3 '13 at 19:29