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Independent statistics: Basketball?

  1. Feb 16, 2009 #1
    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):

    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!
  2. jcsd
  3. Feb 16, 2009 #2
    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 .
  4. Feb 16, 2009 #3
    Thanks much, I actually managed to figure it out while taking a dinner break from problem sets - you beat me back it seems!
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