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Math Help - Need Urgent Help!! Probability Quesiton

  1. #1
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    Need Urgent Help!! Probability Quesiton

    p=80% (chances of winning a game)
    n=5 (number of games)
    expected value of x=4

    How do I find the variance and standard deviation of x?

    And also, how do I find the probability that my opponent will win at least one game


    Thanks!
    Last edited by mr fantastic; November 30th 2009 at 03:49 AM. Reason: Restored original post
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  2. #2
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    Response

    p=80% (chances of winning a game)
    n=5 (number of games)
    expected value of x=4

    How do I find the variance and standard deviation of x?
    well, variance is found by using V = n*p*(1-p)
    V = 5*.80*.20 = 0.80

    The standard deviation is found by taking the square root of the variance.
    sigma = sqrt(V) = sqrt(0.80) = .8944

    I'm gonna try to figure out the second part of the question now.
    Last edited by CaptainBlack; November 8th 2008 at 08:42 AM.
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  3. #3
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    Response

    p=80% (chances of winning a game)
    n=5 (number of games)
    expected value of x=4

    How do I find the variance and standard deviation of x?

    And also, how do I find the probability that my opponent will win at least one game

    I think you would model your opponent winning at least one game as the probability of you winning at most 4 games:

    P total = P(win 0) + P(win 1) + P(win 2) + P(win 3) + P(win4)

    <br />
P(x\!=\!0) \;=\;{5\choose0}(0.8)^0(0.2)^5 \;=\;0.0003<br />

    <br />
P(x\!=\!1) \;=\;{5\choose1}(0.8)^1(0.2)^4 \;=\;0.0064<br />

    <br />
P(x\!=\!2) \;=\;{5\choose2}(0.8)^2(0.2)^3 \;=\;0.0512<br />

    <br />
P(x\!=\!3) \;=\;{5\choose3}(0.8)^3(0.2)^2 \;=\;0.2048<br />

    <br />
P(x\!=\!4) \;=\;{5\choose4}(0.8)^4(0.2)^1 \;=\;0.4096<br />

    Add all of these values to get: P total = 0.6723
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  4. #4
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    Quote Originally Posted by JohhnySD View Post
    solved! Thanks!
    Do not delete questions that you think have been solved. Other members might want to look at the question. And it might not be as solved as you think ....
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  5. #5
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    Quote Originally Posted by ajj86 View Post
    p=80% (chances of winning a game)
    n=5 (number of games)
    expected value of x=4

    How do I find the variance and standard deviation of x?

    well, variance is found by using V = n*p*(1-p)
    V = 5*.80*.20 = 0.80

    The standard deviation is found by taking the square root of the variance.
    sigma = sqrt(V) = sqrt(0.80) = .8944

    I'm gonna try to figure out the second part of the question now.
    Please quote what you are responding to, so we do not have the present situation where we do not know what the original question was because the OP has deleted it.

    Sorry I see that you have quoted it (if you use the quote button rather than the reply it will put the quoted text in a nice box).

    Thanks

    CB
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