# Math Help - Probability equation

1. ## 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

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

3. 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?

4. Originally Posted by Mathematicsfan
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?