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Homework Help: Exam tomorrow - Probability

  1. May 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A football team wants to have a 0,5 chance of scoring on at least 4 out of 5 penalties. What must be the probability p of scoring on each penalty?

    2. Relevant equations
    [tex]{n \choose k} \cdot p^k \cdot (1-p)^{n-k}[/tex]

    3. The attempt at a solution
    Tried this:
    [tex]{5 \choose 4} \cdot p^4 \cdot (1-p)^{1} \ + \ {5 \choose 5} \cdot p^5 \cdot (1-p)^{0} = 0,5[/tex]

    [tex]5 \cdot p^4 \cdot (1-p) + 1 \cdot p^5 \cdot 1 = 0,5[/tex]

    [tex]5p^4 - 5p^5 + p^5 = 0,5[/tex]

    [tex]5p^4 - 4p^5 = 0,5[/tex]

    I don't know what to do with this equation, although someone told me about Newton's approximation something. That is way above the level of this math course anyway, so I guess there is a much easier way to solve this problem. Kinda embarrassing, really, as I take a higher course but came across this problem when helping a younger student.
    Hoping for all the help I can get. :-) Thanks a lot!

    Damn, I don't know what's up with the Latex, but never mind that long, stupid string on the top. Apparently there's something wrong with the spacing, and it won't go away even though I remove it all.
    Last edited: May 21, 2009
  2. jcsd
  3. May 22, 2009 #2


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    Hi Norway! :smile:

    yes, the only way to solve that would be by approximation, or by trial-and-error with a calculator. :wink:
  4. May 22, 2009 #3
    Hm. Is there any alternate way to solve the original problem? I'm absolutely sure that those in this course aren't expected to approximate this, and I don't think they're supposed to sit and try and fail with a calc either.

    I asked this on a Norwegian math forum too, and I was told that my equation was wrong. Well - by inserting the answer given, p=0,686, the equation is correct, but I was told:

    Your equation is wrong. Correct way to solve it:
    [tex]{5 \choose 4} \cdot p^4 \cdot p^1 \geq 0,5[/tex]

    I don't quite understand why this inequality should give the right answer - if it does at all. Does anybody bother to explain?

    Thanks a lot.
  5. May 22, 2009 #4


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    Hi there Norway.

    The best way to think about this is to picture a state with two entries. On and off or win and no win.

    Now the best way to think about this problem is to picture multiple independent events.

    So we start off with representing a state with Win = 0.5 and Lose = 0.5.

    We then move to discussing how we can find the probability of getting 4 events that are
    a win and one that is a loss.

    If we have four out of five wins then we will have the following configuration of events


    So we have five possible combinations of winning four out of five times. This is denoted as 5C4 = 5 ways of winning four out of five times.

    For independent events we multiply the probabilities together. This is based on the multiplication rule whereby multiplying probabilities together is ok because each event is independent of the next. If you learn further statistics you will find that in some cases events aren't independent but in this case if you think about it closely you will realize that each win is independent of every other win or loss.

    So we have 5c4 ways of representing five wins and we have 0.5^4*0.5 probability using the multiplication rule.

    Therefore the total probability is

    P = 5C4 * (0.5)^4*(0.5)

    I think that most of it should be ok to understand but they do not teach properly in high school and it usually takes a while until you get to university to learn this stuff properly.

    Hope that helps.
  6. May 22, 2009 #5


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    Nor do I! :rolleyes:

    There seems no sense behind it at all.
  7. May 22, 2009 #6
    Thanks a lot for your replies!

    I do understand the logic behind nCk and binomial probability etc., and that's how I got to my equation, which seems correct. However, because of the level of this math course vs the level of solving this equation, I believe there must be an easier way.

    I follow you all the way until here:
    Why do you insert 0,5 here?
    The total probability, i.e. the probability for scoring at least 4 out of 5 goals, is 0,5. That means that the probability p for scoring on each penalty must be a lot bigger, and I get the equation:
    [tex]5p^4 - 4p^5 = 0,5[/tex]

    Sorry if I'm being difficult, I just want to understand what's going on here - and if there is an easier way - because as I said, this is a fairly easy math course, intended for young students that are moving on to study languages, law, sociology and other subjects like that (not maths).
  8. May 22, 2009 #7


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    [tex]5p^4 - 4p^5 = 0.5[/tex] seems correct to me as well - I can't argue with your reasoning from the first post. And in fact I can't see how one could justify using the "correct" inequality you were given when it gives the wrong answer! (I wrote a little computer simulator to check that .686 is the correct result)
  9. May 22, 2009 #8


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    I assumed a probability of a loss being 0.5 because a win was 0.5. Since there are only two events we use the probability of the complement of the event is equal to P(Complement(A)) = 1 - P(A). So basically if an event says E = Winning event which is P(E) = 0.5 then a loss must be 1 - 0.5. You can do this by drawing Venn diagrams and making the univeral set equal to "1" or the "whole probability space". I would just ask your tutor about this and i'm sure they can explain it to you.

    Hope that helps.
  10. May 22, 2009 #9


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    I don't think it was mentioned anywhere that the probability of a win is 0.5. I'm assuming you treat each goal scored on a penalty kick as a 'win'. They denoted that as an unknown 'p' instead.
  11. May 22, 2009 #10
    Yup, Defennder, that's what I was thinking too. The guy with the inequality apologized deeply, by the way, and admitted that he was very wrong there. I'm still confused, though, because as I've said numerous times now, this course has just barely introduced binomial probability, and they are currently learning about solving second degree polynomials. There is simply no way they are supposed to solve that fifth degree equation. There has to be a trick to solve it easier than to set up my fifth degree equation! - Or?
  12. May 22, 2009 #11


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    Not only are they not supposed to solve a quintic equation, most quintic equations are not solvable analytically, this one included. Are you allowed to use a graphical calculator during the exam? If so I would wager you are meant to use it to find the value for p numerically.
    To clarify there exists no analytical solution to your problem regardless of the road you take so either you're supposed to not solve it at all or do it graphically/numerically.
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