Answer

From Bayes' theorem

\[ \begin{aligned} f_{Q|X}(q|x) &= \frac{f_{X|Q}(x|q) f_{Q}(q)}{f_{X}(x)}\newline &= \frac{f_{X|Q}(x|q) f_{Q}(q)}{\int_{0}^{1} f_{X|Q}(x|q) f_{Q}(q) dq} \end{aligned} \]

We will need to solve separately for \(x = 0\) and \(x = 1\) as \(x\) is discrete.

\[ \begin{aligned} f_{Q|X=0}(q|x=0) &= \frac{(1-q) \times 6q(1-q)}{\int_{0}^{1} (1-q)\times 6q(1-q) dq} = 12q(1-q)^{2}\newline f_{Q|X=1}(q|x=1) &= \frac{q \times 6q(1-q)}{\int_{0}^{1} q\times 6q(1-q) dq} = 12q^{2}(1-q) \end{aligned} \]