Problem source

My two friends, who shall remain nameless, but whom I shall refer to as $P$ and $Q$, both told me this afternoon that there is a body in my fridge. I'm not sure what to make of this, because $P$ tells the truth with a probability of only $p$, while $Q$ (independently) tells the truth with probability $q$. I haven't looked in the fridge for some time, so if you had asked me this morning, I would have said
that there was just as likely to be a body in it as not. Clearly, in view of what $P$ and $Q$ told me, I must revise this estimate.
Explain carefully why my new estimate of the probability of there being a body in the fridge should be 
\[
\frac{pq}{1-p-q+2pq}.
\]
I have now been to look in the fridge, and there is indeed a body in it; perhaps more than one. It seems to me that only my enemy $A$, or my enemy $B$, or (with a bit of luck) both $A$ and $B$ could be in my fridge, and this morning I would have judged these three possibilities to be equally likely. But tonight I asked $P$ and $Q$ separately whether or not $A$ was in the fridge, and they each said that he was. What should be my new estimate of the probability that both $A$ and $B$ are in my fridge?
Of course, I tell the truth always.

Solution source

\begin{align*}
\mathbb{P}(\text{body in fridge} | \text{P and Q say so}) &= \frac{\mathbb{P}(\text{body in fridge and P and Q say so})}{\mathbb{P}(\text{P and Q say so})} \\
&= \frac{\frac12 pq}{\mathbb{P}(\text{body in fridge and P and Q say so})+\mathbb{P}(\text{no body in fridge and P and Q say so})} \\
&= \frac{\frac12 pq}{\frac12 pq + \frac12(1-p)(1-q)} \\
&= \frac{pq}{pq + 1-p-q+pq} \\
&= \frac{pq}{1-p-q+2pq}
\end{align*}

\begin{align*}
\mathbb{P}(\text{A and B in fridge} | \text{P and Q say A is in fridge}) &= \frac{\mathbb{P}(\text{A and B in fridge and P and Q say A is in fridge}) }{\mathbb{P}( \text{P and Q say A is in fridge}) } \\
&= \frac{\frac13pq}{\frac13pq+\frac13pq+\frac13(1-p)(1-q)} \\
&= \frac{pq}{1-p-q+3pq}
\end{align*}