# Events A and B are independent such that $P(A)=6P(B)$

Events A and B are independent such that $P(A)=6P(B)$ and $P(A \cup B) =0.915.$ Find P(B).

I know that $P(A \cup B)= P(A) + P(B) -P(A)P(B)$

Then $0.915=6P(B)+P(B)-6P(B)P(B) \\ \rightarrow 0.915=7P(B)-6P(B)^2$.

$-6P(B)^2+7P(B)-0.915=0$ Letting $a=-6, b=7, \& c=-0.915$

I can use the quadratic formula: $\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$ Which gives me the roots of $1.0166..$ and $0.15$, since the $P(B) < 1$ then $P(B)$ is $0.15?$

Am I taking this the correct direction?

• yes, you are. Just finish the quadratic equation – Arnaldo Feb 9 '17 at 20:22
• Okay that was just the direction I needed to take, thanks! – Killercamin Feb 9 '17 at 20:33
• You did it well! – Arnaldo Feb 9 '17 at 20:36