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1987 Paper 3 Q15
D: 1500.0 B: 1500.0

A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.

  1. Show that \[ E_{n+1}-E_{n}=10p-10(1-p)q^{n}(9-10q), \] and deduce that if \(p=10^{-4},q=10^{-2},\) she should choose \(n=3.\)
  2. Show that if \(q\) is larger than some fixed value \(q_{0},\) to be determined explicitly, then whatever the value of \(p,\) she should choose \(n=0.\)


Solution:

  1. \(E_{n+1} - E_n\) is the additional cost of the extra test \(10p+10(1-p)q^n\) from people who have yet to fail a test plus the reduce cost of people who will fail the final test, \(-100(1-p)q^n(1-q)\) \begin{align*} E_{n+1}-E_{n} &= 10p+10(1-p)q^n-100(1-p)q^n(1-q) \\ &=10p +10(1-p)q^n(1-10(1-q)) \\ &= 10p +10(1-p)q^n(-9+10q) \\ &= 10p - 10(1-p)q^n(9-10q) \end{align*} \begin{align*} && 10p - 10(1-p)q^n(9-10q) &> 0 \\ \Leftrightarrow && \frac{p}{(1-p)(9-10q)} &>q^n \end{align*} If \(p = 10^{-4}, q = 10^{-2}\) we have: \begin{align*} \frac{p}{(1-p)(9-10q)} &= \frac{10^{-4}}{(1-10^{-4})(9-10^{-3})} \\ &\approx 10^{-5} \end{align*} and \(q^2 < 10^{-5} < q^3\) so she should stop after the 3rd test.
  2. She shouldn't bother testing if \begin{align*} && \frac{p}{(1-p)(9-10q)} &>1 \\ \Leftrightarrow && \frac{p}{1-p} &>9-10q \\ \Leftrightarrow && 10q &>9 \\ \Leftrightarrow && q &> \frac9{10} = q_0 \end{align*}