1. Suppose that n people take a blood test for a disease, where each person has probability p of having
the disease, independent of other people. To save time and money, blood samples from k people are
pooled and analyzed together. If none of the k persons has the disease then the test will be negative,
but otherwise it will be positive. If the pooled test is positive then each of the k persons is tested
separately (so k + 1 tests are done in that case).
[3] (a) Let X be the number of tests required for a group of k people, then X takes possible values at 1
or k + 1, depending on the outcome of the pooled test. Show that
k
E(X) = k + 1 ▯ k(1 ▯ p) :
Soln: f(1) = P(X = 1) = (1 ▯ p) and f(k + 1) = P(X = k + 1) = 1 ▯ (1 ▯ p) . k
E(X) = 1 ▯ f(1) + (k + 1)f(k + 1) = (1 ▯ p) + (k + 1)[1 ▯ (1 ▯ p) ] = k + 1 ▯ k(1 ▯ p) : k
[2] (b) What is the expected number of tests required for n=k groups of k people each? (Note: the total
number of people involved is n)
Soln: The expected number of tests is (n=k)E(X) = n + n=k ▯ n(1 ▯ p) k
▯1
[1] (c) Show that if p is small, the expected number of tests in part (b) is approximately n(kp + k ),
and is minimized for k = p ▯1=2.
k 2 k :
Soln: Noting that (1▯p) = 1▯kp+k(k +1)p +▯▯▯, we have (1▯p) = 1▯kp when p is small.
So
k : ▯1
n + n=k ▯ n(1 ▯ p) = n + n=k ▯ n(1 ▯ kp) = n(kp + k ):
2 2
Using the fact that a + b ▯ 2ab, and the equality holds when a = b, we have
▯1 q p p
n(kp + k ) ▯ 2n kp ▯ k▯1 = 2n p;
p ▯1 ▯1=2
and the minimum value 2n p is achieved when kp = k , i.e., k = p . 2. Let X be a random variable taking values in the set f0;1;2g, with moments ▯ = E(X) = 1 and 1
2
▯ 2 E(X ) = 3=2.
[2] (a) Let p = i(X = i), i = 0;1;2. Show that p = 1=4, p = 1=0 and p = 1=1. 2
2
Soln: From 1 = E(X) = p + 2p and 1=2 = E2X ) = p + 4p , we

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