1 problem found
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: \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*}