Suppose we have a population of N people, each carries a certain virus with prob

Question

Suppose we have a population of N people, each carries a certain virus with probability p, and these are independent events. We have a blood test that detects the virus perfectly. We can test each individual separately, but that will involve N tests. Alternately, we can pool k samples together, and then, if the test is negative, we can declare all the corresponding k people virus-free, or, if we get a positive answer, we can test each of those k individually, for a total of k+1 tests for that group. What is the expected number of tests necessary under the alternate procedure? (Assume that k divides N.) If p is small, what values of k would you choose to minimize the number of tests? Thanks in advance for the help! I appreciate any help and I will review positively.

Explanation / Answer

For each group, the probability for each person in the group to have the virus is p. Since these are independent events, we can calculate the probability of a negative group as: (1-p)^k (since each member of the group must be negative) That means that the probability of having a positive group is: 1- (1-p)^k Now a positive group means that we have to administer k+1 tests and a negative group means that we only have to administer 1 test. So the expected value for each group is: (1-(1-p)^k)(k+1) + ((1-p)^k)(1) = (1-(1-p)^k)(k+1) + (1-p)^k = k-(k+1)(1-p)^k+1+(1-p)^k So, this simplifies to: k - k(1-p)^k + 1 Now, the number of groups is N/k, so the total expected value of tests is: (N/k)(k - k(1-p)^k + 1) = N - N(1-p)^k + N/k So, the smaller p gets, the bigger we we want to make k. As p-->0, k--> N

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