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Math Help - Probability

  1. #1
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    Exclamation Probability

    A TV sports commentator claims that 45 percent of all football injuries are knee injuries, assuming that this claim is true what is the probability that in a game with 5 reported injuries:
    a) all are knee injuries?
    b) none are knee injuries?
    c) at least 3 are knee injuries?
    d) no more than 2 are knee injuries?

    I am looking for some direction, confused as to where to begin, looking for any kind of help, it'd be much appreciated.
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  2. #2
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    Quote Originally Posted by cvaughn13 View Post
    A TV sports commentator claims that 45 percent of all football injuries are knee injuries, assuming that this claim is true what is the probability that in a game with 5 reported injuries:
    a) all are knee injuries?
    b) none are knee injuries?
    c) at least 3 are knee injuries?
    d) no more than 2 are knee injuries?

    I am looking for some direction, confused as to where to begin, looking for any kind of help, it'd be much appreciated.
    a) P("all are knee injuries") = 0.45*0.45*0.45*0.45*0.45 = 0.45^5

    b) P("none are knee injuries") = (1-0.45)^5 = 0.55^5,

    because you know P("knee injury") = 0.45
    then P("no knee injury") = 1-0.45 = 0.55

    c) P("at least 3 knee injuries") = \begin{pmatrix} 5 \\ 3\end{pmatrix}*0.45^3*0.55^2 + \begin{pmatrix} 5 \\ 4\end{pmatrix}*0.45^4*0.55^1+0.45^5

    I consider "A and B and C has a knee injury, but D and E don't and D and E and A has a knee injury, but player B,C don't" as two different cases. Thats where the 5 over 3 comes from.
    Last edited by Rapha; May 29th 2009 at 12:39 AM. Reason: some typing mistakes
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  3. #3
    MHF Contributor matheagle's Avatar
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    Sorry, found some errors, see next post.
    This is binomial with n=5 and p=P(knee injury)=.45
    (d) is P(0 or 1 knee injury)=P(X=0)+P(X=1)
    Last edited by matheagle; May 28th 2009 at 09:53 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    There's a typo in (c) that I just noticed. I'm changing the 4 to a 1 here, above the q.

    AND (b) is partially right/wrong.

    The answer is .55^5 which is not equal to  1-0.45^5

    Maybe you meant  (1-0.45)^5 , which is okay, but that's not how I would interprete your math.

    Quote Originally Posted by Rapha View Post
    a) P("all are knee injuries") = 0.45*0.45*0.45*0.45*0.45 = 0.45^5

    b) P("none are knee injuries") = 1-0.45^5 = 0.55^5,

    because you know P("knee injury") = 0.45
    then P("no knee injury") = 1-0.45 = 0.55

    c) P("at least 3 knee injuries") = \begin{pmatrix} 5 \\ 3\end{pmatrix}*0.45^3*0.55^2 + \begin{pmatrix} 5 \\ 4\end{pmatrix}*0.45^4*0.55^1+0.45^5

    I consider "A and B and C has a knee injury, but D and E don't and D and E and A has a knee injury, but player B,C don't" as two different cases. Thats where the 5 over 3 comes from.
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