# Problem 352: Blood tests
Each one of the 25 sheep in a flock must be tested for a rare virus,
known to affect 2% of the sheep population. An accurate and extremely
sensitive PCR test exists for blood samples, producing a clear positive
/ negative result, but it is very time-consuming and expensive. Because
of the high cost, the vet-in-charge suggests that instead of performing
25 separate tests, the following procedure can be used instead: The
sheep are split into 5 groups of 5 sheep in each group. For each group,
the 5 samples are mixed together and a single test is performed. Then,
If the result is negative, all the sheep in that group are deemed to be
virus-free. If the result is positive, 5 additional tests will be
performed (a separate test for each animal) to determine the affected
individual(s). Since the probability of infection for any specific
animal is only 0.02, the first test (on the pooled samples) for each
group will be: Negative (and no more tests needed) with probability
0.985 = 0.9039207968. Positive (5 additional tests needed) with
probability 1 - 0.9039207968 = 0.0960792032. Thus, the expected number
of tests for each group is 1 + 0.0960792032 × 5 = 1.480396016.
Consequently, all 5 groups can be screened using an average of only
1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of
more than 70% ! Although the scheme we have just described seems to be
very efficient, it can still be improved considerably (always assuming
that the test is sufficiently sensitive and that there are no adverse
effects caused by mixing different samples). E.g.: We may start by
running a test on a mixture of all the 25 samples. It can be verified
that in about 60.35% of the cases this test will be negative, thus no
more tests will be needed. Further testing will only be required for the
remaining 39.65% of the cases. If we know that at least one animal in a
group of 5 is infected and the first 4 individual tests come out
negative, there is no need to run a test on the fifth animal (we know
that it must be infected). We can try a different number of groups /
different number of animals in each group, adjusting those numbers at
each level so that the total expected number of tests will be minimised.
To simplify the very wide range of possibilities, there is one
restriction we place when devising the most cost-efficient testing
scheme: whenever we start with a mixed sample, all the sheep
contributing to that sample must be fully screened (i.e. a verdict of
infected / virus-free must be reached for all of them) before we start
examining any other animals. For the current example, it turns out that
the most cost-efficient testing scheme (we'll call it the optimal
strategy) requires an average of just 4.155452 tests! Using the optimal
strategy, let T(s,p) represent the average number of tests needed to
screen a flock of s sheep for a virus having probability p to be present
in any individual. Thus, rounded to six decimal places, T(25, 0.02) =
4.155452 and T(25, 0.10) = 12.702124. Find ΣT(10000, p) for p=0.01,
0.02, 0.03, ... 0.50. Give your answer rounded to six decimal places.