Given $Pr(A) = 0.3, Pr(B) = 0.4$, and $Pr(A'|B') = \frac{10p^2-1}{6}$, find the possible values of $p$.
From what I know(using the conditional probability formula): $$Pr(A'|B')= \frac{Pr(A'\cap B')}{Pr(B')}$$
$$Pr(A'\cap B') = \frac{10p^2 -1}{10}$$
I need help after this step.
$Pr(A' \cap B')$=$1-Pr(A \cup B)$=$1-[0.4+0.3-Pr(A \cap B)]$
=$0.3+Pr(A \cap B)$
Hence on simplification, we get $p^2=0.4+Pr(A \cap B)$.
Also, $0<Pr(A \cap B)<min[Pr(A),Pr(B)]=0.3$
Hence, $0.4<p^2<0.7$. So, you get all possible values of $p^2$, and hence $p$.