Independent statistics: Basketball?

[evildaemonlad]

Hi, I'm a bit stuck on this problem

A basketball player misses 30% of his free throws. He ends up in a situation where he has the potential to shoot two penalty shots if and only if he lands the first shot (called a one and one, I believe). The outcome of the 2nd shot is independent of the first.

> Find the expected number of points resulting from the "one and one"

Here I'm assuming expected number equates to expected value -> E(x) = S x P(x) (S = sigma)

I've set up a table thusly so that I might take the sum of the 3rd column (x denotes successful shots):
x____P(x)____xP(x)
0...?...0
1.....
2.....

The issue that I am having is how to express the dependence of even taking a second shot upon this first. My gut is telling me that the first entry under P(x) ought to be .3 but after that I'm lost. Any pointers would be appreciated!

 

[Rogerio]

Then, 30% of the "one and one" , he misses the first shot, and gets 0 points (prob 30%).
And he lands 70% of the first shot.
Then, he misses the second throw, getting only 1 point (prob 0.7 * 0.3 = 21%).
But he lands 70% of the 2nd throw, getting 2 points (prob 0.7 * 0.7 = 49%).

So, the expected number of points is 1*0.21 + 2*.49 = 1.19 .

 

[evildaemonlad]

Thanks much, I actually managed to figure it out while taking a dinner break from problem sets - you beat me back it seems!