Problem 70: Bodies in the fridge ($\u2713$ $\u2713$) 1987 Paper II

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 $p$, while $Q$ (independently) tells the truth with a probability of only $q$. 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 the fridge as not. Clearly, in view of what my friends have 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 ${E}_{1}$ or my other enemy ${E}_{2}$ or (with a bit of luck) both ${E}_{1}$ and ${E}_{2}$ could be in my fridge, and this evening I would have judged these three possibilities equally likely. But tonight I asked $P$ and $Q$ separately whether ${E}_{1}$ was in the fridge, and they each said that she was. What should be my new estimate of the probability that both ${E}_{1}$ and ${E}_{2}$ are in my fridge?

Of course, I always tell the truth.

Comments

The most difficult part of this problem is unravelling the narrative! The ﬁrst paragraph says essentially ‘what is the probability that there is body in the fridge, given that $P$ and $Q$ both say there is?’. It can therefore be tackled by the usual methods of conditional probability: tree diagrams, for example, or Bayes’ theorem. All the other words in the ﬁrst paragraph are there to tell you about the a priori probabilities of the events, without knowledge of which the question above would be meaningless.

In the second paragraph, the situation becomes more complicated, but the method used for the ﬁrst paragraph will still work.

In case you want to use it, here is the statement of Bayes’ theorem, in its simplest form:

$$\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(B\right)\times \mathrm{P}\left(A|B\right)}{\mathrm{P}\left(A\right)}.$$

Solution to problem 70

This problem can be solved using tree diagrams. A more sophisticated, but not necessarily better, method is to use Bayes’ theorem.

Here, we take the events $A$ and $B$ to be

$$\begin{array}{rcl}A& =& \text{}P\text{and}Q\text{bothsaythatthereisabodyinthefridge}\\ B& =& \text{thereisabodyinthefridge}\\ & \end{array}$$From the information given in the question, $P\left(B\right)=\frac{1}{2}$, so

$$\mathrm{P}\left(B|A\right)=\frac{\frac{1}{2}\times pq}{\mathrm{P}\left(A\right)}.$$Now

$$\begin{array}{rcll}\mathrm{P}\left(A\right)& =& \mathrm{P}\left(\text{thereisabody}\right)\times \mathrm{P}\left(\text{}P\text{and}Q\text{bothsaythereis}\right)& \text{}\\ & & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\mathrm{P}\left(\text{thereisnotabody}\right)\times \mathrm{P}\left(\text{theybothsaythereis}\right)& \text{}\\ & =& \frac{1}{2}\times pq+\frac{1}{2}\times \left(1-p\right)\left(1-q\right)& \text{}\end{array}$$which gives the required answer.

For the second paragraph, let

$$\begin{array}{rcll}X& =& \text{}P\text{and}Q\text{bothsaythat}{E}_{1}\text{isinthefridge}& \text{}\\ Y& =& \text{}{E}_{1}\text{and}{E}_{2}\text{areinthefridge}& \text{}\end{array}$$There are three possibilities (since we know that there is at least one body in the fridge): only ${E}_{1}$ is in the fridge; only ${E}_{2}$ is in the fridge; and both ${E}_{1}$ and ${E}_{2}$ are in the fridge. These are given as equally likely, so the a priori probabilities are each $\frac{1}{3}$.

We want $\mathrm{P}\left(Y|X\right)$, which by Bayes’ theorem is

$$\frac{\mathrm{P}\left(Y\right)\times \mathrm{P}\left(X|Y\right)}{\mathrm{P}\left(X\right)}=\frac{\frac{1}{3}\times pq}{\mathrm{P}\left(X\right)}.$$Now

$$\begin{array}{rcll}\mathrm{P}\left(X\right)& =& \mathrm{P}\left(\text{only}{E}_{1}\text{isinthefridge}\right)\times \mathrm{P}\left(\text{}P\text{and}Q\text{toldthetruth}\right)& \text{}\\ & & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\mathrm{P}\left(\text{both}{E}_{1}\text{and}{E}_{2}\text{areinthefridge}\right)\times \mathrm{P}\left(\text{}P\text{and}Q\text{bothtoldthetruth}\right)& \text{}\\ & & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}+\mathrm{P}\left(\text{only}{E}_{2}\text{isinthefridge}\right)\times \mathrm{P}\left(\text{}P\text{and}Q\text{bothlied}\right)& \text{}\\ & =& \frac{1}{3}\times pq+\frac{1}{3}\times pq+\frac{1}{3}\times \left(1-p\right)\left(1-q\right)\\ & \end{array}$$so the answer is $\frac{pq}{1-p-q+3pq}$.

Post-mortem

Note how much more difficult it is when the answer is not given; when the question was originally set, most candidates arrived at the given answer to the ﬁrst part but were not sufficiently conﬁdent to extend their method to the second paragraph: they received $\frac{8}{20}$ for their efforts.

The last line of the question is not entirely frivolous; if I may have lied about what my friends answered when I asked them if there is a body in the fridge, the problem becomes difficult. However, my claim to be truthful is vacuous (it tells you nothing) because I may be lying. Contrast with the statement ‘I am lying’, which is inconsistent.