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Math Help - Probability equation

  1. #1
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    Probability equation


    I received these two equations and my goal is to find variable P
    It should be done by dividing two equation together.
    the problem is that I'm stuck I don't how to start it.


    final result : 0.3
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  2. #2
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    Hello, Mathematicsfan!

    The equations are strangely written
    . . and I don't agree with their answer.


    \begin{array}{ccc}6p^2(1-p)^2 \;=\;0.0016 & [1] \\ p^4 \;=\;0.0625 & [2] \end{array}

    Divide [1] by [2]: . \dfrac{6p^2(1-p)^2}{p^4} \;=\;\dfrac{0.0016}{0.0625} \quad\Rightarrow\quad \dfrac{6(1-p)^2}{p^2} \;=\;0.0256

    . . 6(1-p)^2 \:=\:0.0256p^2 \quad\Rightarrow\quad 6 - 12p + 6p^2 \:=\:0.0256p^2

    . . 5.9744p^2 - 12p + 6 \;=\;0


    Quadratic Formula: . p \;=\;\dfrac{12 \pm \sqrt{12^2 - 4(5.9744)(6)}}{2(5.9744)} \;=\;\dfrac{12 \pm \sqrt{0.6144}}{11.9488}


    Therefore: . p \;=\;\begin{Bmatrix}1.069\,884\,567 \\ 0.938\,685\,331 \end{Bmatrix}

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  3. #3
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    Perhaps I made a miscalculations ?
    here is the original task:

    Team A and Team B play between themselves 4 games of football.
    It's known that the probability that team A will win all 4 games is 0.0625 and the probability that all 4 games will end with a draw (2 out of 2 events) is 0.0016.
    What's the probability that team B will win in single game?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Mathematicsfan View Post
    Perhaps I made a miscalculations ?
    here is the original task:

    Team A and Team B play between themselves 4 games of football.
    It's known that the probability that team A will win all 4 games is 0.0625 and the probability that all 4 games will end with a draw (2 out of 2 events) is 0.0016.
    What's the probability that team B will win in single game?
    {p_A}^4=0.0625

    {p_A}^2{p_B}^2=0.0016

    Can you proceed?
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