Free Throw Probability and Expected Number of Points 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.
 A: 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,

\begin{align}E(P)&=E(T_1)+E(T_2)\\&=0.8+0.8=1.6\\Var(P)&=Var(T_1+T_2)\\&=Var(T_1)+Var(T_2)\\&=0.8\times(1-0.8)+0.8\times(1-0.8)\end{align}

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}

A: 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.  
