A basketball player who shoots 80% from the free-throw line goes to the charity stripe with just 1.7 seconds remaining in a basketball game.
He has two shots, and his team is trailing by two points.

What is the probability that he will make only one of the two, and why?

(Assume that he is trying, of course, for his team to win the game.)

There are only 4 possible outcomes:
1. Make both - prob .8 x .8 = .64
2. Miss both - prob = .2 x .2 = .04
3. Make first, miss second - prob = .8 x .2 = .16
4. Miss first, make second - prob = .2 x .8 = .16

You can just add #s 3 and 4, OR you could've done 1 - (#1 +#2) = 1 - (.64 + .04) = .32

Hope I get this one, unlike the gears. Don't think half a rotation on the bball matters here, though.

Posted by Lawrence on 2003-08-24 06:18:54