Suppose any child is male with probability $p$ or female with probability $1 − p$, independently of other children. In a family with four children, let $A$ be the event that there is at most one girl, and $B$ the event that there are children of both sexes. Show that there is a value of $p$, with $0 < p < 1/2$, such that $A$ and $B$ are independent.
MY ATTEMPT
In the first place, we get $\textbf{P}(A) = p^{4} + 4p^{3}(1-p)$ and $\textbf{P}(B) = 1 - p^{4} - (1-p)^{4}$. Finally, we have $\textbf{P}(A\cap B) = 4p^{3}(1-p)$. According to the definition of independency, we obtain the following equation to solve
\begin{align*} 4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}] \end{align*}
Wolfram gives $1$ and $0.41133$ as the reals solution, but I suspect there is another approach to the problem. Could someone provide me an answer or an alternative way to tackle this exercise? I will be grateful anyway. Thanks in advance.
Use computer algebra to get all the answers:
$$\left\{0,0,0,1,1,\frac{1}{9} \left(4-\frac{38}{\sqrt[3]{9 \sqrt{681}-17}}+\sqrt[3]{9 \sqrt{681}-17}\right),\frac{4}{9}+\frac{19 \left(1+i \sqrt{3}\right)}{9 \sqrt[3]{9 \sqrt{681}-17}}-\frac{1}{18} \left(1-i \sqrt{3}\right) \sqrt[3]{9 \sqrt{681}-17},\frac{4}{9}+\frac{19 \left(1-i \sqrt{3}\right)}{9 \sqrt[3]{9 \sqrt{681}-17}}-\frac{1}{18} \left(1+i \sqrt{3}\right) \sqrt[3]{9 \sqrt{681}-17}\right\}$$
Why does this not suffice?