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Suppose a basketball player has an 80 percent chance of making a free throw. He has been fouled, and has two free throws. If a free throw is made, its counts as one point. Let $X$ and $Y$ be the number of points from the first and second free throw respectively. Figure out the joint probabilities for $X$ and $Y$, the expected number of points from the two free throws, and the variance for this number of points, when:

a) each free throw is an independent event

b) $P(Y=1\mid X=1)=0.9$ and $P(Y=1\mid X=0)=0.4$ so that making the first free throw raises the probability of making the second free throw.

So I've pretty much done part a but the only thing I'm confused about is when I find the expected number of points from the two free throws its less than 1. Should it be that way? Than, for part b it only gives me two of the probabilities and I'm a little confused about how to find the other two. I tried setting it up into a two by two box with X values on top and Y values on the side but still couldn't figure that out. Any input would be appreciated. Thanks!

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marked as duplicate by apnorton, AlexR, amWhy, ᴡᴏʀᴅs, RecklessReckoner Jul 17 '14 at 17:40

This question was marked as an exact duplicate of an existing question.

Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Figure out..."), not a request for help, so consider rewriting it. – Zev Chonoles Feb 4 '13 at 4:37
Thanks! The "figure out" was just part of the problem itself so I wasn't trying to be rude or anything. I'll keep all that in mind and I appreciate it. – Gamecocks99 Feb 4 '13 at 4:39
Another user asked this question one day later, but the other question has a more detailed answer. (FYI for future visitors) – apnorton Jul 17 '14 at 15:41

HINT: Find out the results of the 4 cases first:

$$\text {with} \space P(Y=1\mid X=1)=0.9 \space \text{you can find}\space P(Y=0 \mid X=1) = 0.1$$ $$\text {with} \space P(Y=1\mid X=)=0.9 \space \text{you can find}\space P(Y=0 \mid X=1) = 0.1$$

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