Two independent random geometric variables 
Correct Answer = 0.0495
My work:
X, Y~geom(p)
$F(2, 2) = P(1, 1) + P(1, 2) + P(2, 1) + P(2, 2) = p^2 + 2p(1-p) + p^2(1-p)^2 = 0.0441$
I think this is the right step, not sure how to solve equation involving $p^4$..
(Finan Exam P 40.24)

Let $X$ and $Y$ be the number of attempts made by A and B. We can assume these random variables to be independent, so
$$
F(2,2)=\mathbb P(X\leq 2, Y\leq 2) = \mathbb P(X\leq 2)\cdot \mathbb P(Y\leq 2) = (p+p(1-p))^2 =(2p-p^2)^2=(1-(1-p)^2)^2 = 0.0441.
$$
The equality $(1-(1-p)^2)^2 = 0.0441$ follows also directly from the fact that $$\mathbb P(X\leq 2)=1-\mathbb P(X>2)=1-(1-p)^2$$ so you don’t even have to solve the quadratic equation to find $p$.
