[Math] Free Throw Probability and Expected Number of Points

Question

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.

Answer 1

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}