final result : 0.3
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.
Hello, Mathematicsfan!
The equations are strangely written
. . and I don't agree with their answer.
$\displaystyle \begin{array}{ccc}6p^2(1-p)^2 \;=\;0.0016 & [1] \\ p^4 \;=\;0.0625 & [2] \end{array}$
Divide [1] by [2]: .$\displaystyle \dfrac{6p^2(1-p)^2}{p^4} \;=\;\dfrac{0.0016}{0.0625} \quad\Rightarrow\quad \dfrac{6(1-p)^2}{p^2} \;=\;0.0256$
. . $\displaystyle 6(1-p)^2 \:=\:0.0256p^2 \quad\Rightarrow\quad 6 - 12p + 6p^2 \:=\:0.0256p^2 $
. . $\displaystyle 5.9744p^2 - 12p + 6 \;=\;0$
Quadratic Formula: .$\displaystyle p \;=\;\dfrac{12 \pm \sqrt{12^2 - 4(5.9744)(6)}}{2(5.9744)} \;=\;\dfrac{12 \pm \sqrt{0.6144}}{11.9488} $
Therefore: .$\displaystyle p \;=\;\begin{Bmatrix}1.069\,884\,567 \\ 0.938\,685\,331 \end{Bmatrix} $
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?