Conditional Probability - truth/lie

Let $A$ be the event that the first player says the card is a queen, $B$ be the event that the second player says the card is a queen, and $Q$ be the event that a queen is actually drawn. By the problem statement, we can reasonably assume that $A$ and $B$ are independent of each other, when $Q$ is held fixed.

Then we are given $P(Q),P(A|Q),P(B|Q),P(A|Q^C),P(B|Q^C)$, and we are looking to find $P(Q|AB)$.

By Bayes' Theorem, we know:

$$P(Q|AB)=\frac{P(AB|Q)\cdot P(Q)}{P(AB)} \\=\frac{P(AB|Q)\cdot P(Q)}{P(AB|Q)\cdot P(Q)+P(AB|Q^C)\cdot P(Q^C)} \\=\frac{P(A|Q)\cdot P(B|Q)\cdot P(Q)}{P(A|Q)\cdot P(B|Q)\cdot P(Q)+P(A|Q^C)\cdot P(B|Q^C)\cdot P(Q^C)} \\=\frac{(3/5)(2/3)(1/13)}{(3/5)(2/3)(1/13)+(2/5)(1/3)(12/13)} \\=\frac{6}{6+24}=\frac{1}{5}$$