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Suppose a basketball player has an 80% chance of making a free throw. The player was fouled and now has two free throws. If a free throw is made it counts as one point. Let X be the number of points from the first free throw and Y be the number of points from the second free throw. Figure out the joint probabilities from X and Y (I've already done this step), the expected number of points from the two free throws and the variance for this number of points if:

a) Each free throw is an independent event

b) P(Y=1|X=1)=.9 and P(Y=1|X=0)=.4 so that making the first free throw raises the prob. a making the second free throw.

All I really need help on is the expected number of points and the variance for that number of points.

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Why don't you also type out what you have done, it's easier for us to build on from there. Anyways, \begin{align} E(P)&=E(T_1)+E(T_2)\\ Var(P)&=Var(T_1+T_2) \end{align} P is the points and $T_i$ are the throws. If Independent,


If they are not Independent, then:

\begin{align}E(P)&=E(T_1)+E(T_2)\\&=E(T_1)+E(E(T_2)|T_1)\\&=0.8+0.8*0.9+0.2*0.4]\\&=1.6\\Var(P)&=Var(T_1+T_2)\\&=Var(T_1)+Var(T_2)+2\times Cov(T_1,T_2)\\Cov(T_1,T_2) &=E\{(T_1-E(T_1))(T_2-E(T_2))\}\\&=\vdots\\\end{align}

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We do the calculation for (b). The calculation for (a) can be done in a similar way, but also in various nicer ways.

Let $T$ be the number of points. We find the distribution of $T$.

First we find the probability that $T=0$. The probability the first throw is a miss is $0.2$. The probability the second throw is a miss given that the first is a miss is $0.6$. Thus the probability of missing both throws is $(0.2)(0.6)$, which is $0.12$.

Next we could find the probability that $T=1$. But we will be sneaky, and find first the probability that $T=2$. For this we need hit, then hit. The probability the first throw is a hit is $0.8$. Given success on the first trial, the probability of success on the second is $0.9$. So the probability that $T=2$ is $0.72$.

It follows that the probability that $T=1$ is $1-(0.12+0.72)$, which is $0.16$.

We could also have calculated this probability directly, by looking at the two cases hit then miss, and miss then hit.

Now the expectation is easy: $$E(T)=(0)(0.12)+(1)(0.16)+(2)(0.72)=1.60.$$

For the Variance, there are two ways to tackle the problem. Either we calculate $E(T-1.6)^2$, or we use the usually easier formula $\text{Var}(X)=E(X^2)-(E(X))^2$. Let's do it the second way. The random variable $X^2$ is $0$ with probability $0.12$, $1$ with probability $0.16$, and $4$ with probability $0.72$. Thus $$E(X^2)=(0)(0.12)+(1)(0.16)+(4)(0.72).$$ The result is $3.04$. Subtract $(E(X))^2$, that is, $(1.6)^2$. The variance of $T$ is $0.48$.

Another calculation of variance: If we use the defining formula for variance, which in this case is $E(T-1.6)^2$, we get that $$\text{Var}(T)=(-1.6)^2(0.12)+(0.6)^2(0.16)+(0.4)^2(0.72).$$ The calculation is messier than our first calculation, but should give the same answer.

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