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Expected Value (probability problem)

  1. Oct 11, 2009 #1
    Prizes and the chances of winning in a sweepstakes are given in the table below.
    Prize: Chances
    $25,000,000: 1 chance in 400,000,000
    $250,000: 1 chance in 150,000,000
    $75,000: 1 chance in 50,000,000
    $10,000: 1 chance in 4,000,000
    $800,000: 1 chance in 500,000
    A watch valued at $70: 1 chance in 8,000

    Find the expected value (in dollars) of the amount won by one entry.

    E(X) = X_1*P(X_1)+X_2*P(X_2)+....

    3. The attempt at a solution

    I used the formula listed above and do not get the right answer. I'm really unsure as to why that doesn't work. I use the dollar values for my X and then the chance as my P(X). Can someone explain to me where my fault is in working this problem?
  2. jcsd
  3. Oct 11, 2009 #2


    Staff: Mentor

    Please show us your work. It might be that you made an arithmetic mistake that is causing your answer to be incorrect.

    As a side note, it's odd that the probability of winning $25M is relatively high, compared to the probability of winning a prize that is only 1% of that.

    Also, you show 1 chance in 4 million of winning $10,000, for 1 chance in 1/2 million of winning $800,000. Are you sure that you have written all the prizes and their probabilities exactly as presented in the problem?
  4. Oct 11, 2009 #3
    - I copied the values straight from the webwork homework. I do find it very odd that the numbers go from $10,000 @ 1/4,000,000 chance to $800,000 @ 1/500,000 chance. But here is how I worked the problem:

    E(X) = (25,000,000*(1/400,000,000))+(250,000*(1/150,000,000))+(75,000*(1/50,000,000))+(10,000*(1/4,000,000))+(800,000*(1/500,000))+(70*(1/8,000)) = 1.67691666666667
  5. Oct 11, 2009 #4


    Staff: Mentor

    I don't see anything wrong with it. I get the same answer.
  6. Oct 11, 2009 #5
    - yeah, I'm at a loss. Maybe the numbers are showing up wrong in relation to the actual answer.
  7. Oct 11, 2009 #6
    Alright, I figured it out. The $800,000 value is supposed to be $800. So I suppose my professor made a typo when setting the problem up.
  8. Oct 12, 2009 #7


    Staff: Mentor

    If you get the "right" answer when you replace $800K by $800, that probably is what happened.
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