Solution

Suppose there is just one person who has disease.

6 blood tests would be required to detect the diseased person when the diseased person's blood is tested last. The number of ways this can happen is given the number of ways four people can be arranged i.e. $4!.$

The total number of ways of testing the blood is $5!$.

Thus, the probability that six blood tests are required to detect diseased person is $\dfrac{4!}{5!}= \dfrac{1}{5}.$

Alternatively, one could think that all people's blood have equal chance of getting tested at the end and probability of that is $\dfrac{1}{5}.$

Suppose there are two people who have disease.

6 blood tests would be required to detect the diseased person when the one of the diseased person's blood is tested last. The number of ways this can happen is given the number of ways one can choose 2 of the diseased person ( that would be placed at last ) and four ( rest ) people can be arranged i.e. $C_1^24!.$

The total number of ways of testing the blood is $5!$.

Thus, the probability that six blood tests are required to detect both diseased person is $\dfrac{C_1^24!}{5!}= \dfrac{2}{5}.$

Alternatively, one could think that all people's blood have equal chance of getting tested at the end and probability of that is $\dfrac{1}{5}.$ When any of the two diseased person are at the end, we require 6 tests to detect the diseased people. This, gives the probability of such event as 

$\dfrac{1}{5} + \dfrac{1}{5}= \dfrac{2}{5}.$

$Answer$

1/5 = 2/5

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