Let \(U\) be a uniformly distributed random variable on [0,1]. What is the probability that the equation
\[x^2 + 4Ux + 1 = 0\]
has two distinct real roots \(x_1\) and \(x_2\)?
The quadratic equation has real and distinct roots iff its discriminant \(b^2-4ac\) is positive. In this case, the random variable is part of the coefficient \(b\). Specifically, we observe that:
\[ a = 1, b= 4U, c = 1\]
This implies \[b^2 - 4ac = (4U)^2 - 4(1)(1) = 16U^2 - 4 \implies U^2 \geq \frac{1}{4}\]
Since \(U\) is restricted to the interval [0,1], the solution set S of \(U^2 \geq \frac{1}{4}\) is the interval \[ S = \left[\frac{1}{2}, 1\right]\]
Since the probability distribution is uniform, the probability of the solution set \(S\) in \(U\) is proportional to its length inside the [0,1] interval. The probability is therefore 0.5.
(Feller) A large number, N, of people are subjected to a blood test. This can be administered in two ways: (1) Each person can be tested separately, in this case N test are required, (2) the blood samples of k persons can be pooled and analyzed together. If this test is negative, this one test suffices for the k people. If the test is positive, each of the k persons must be tested separately, and in all, k + 1 tests are required for the k people. Assume that the probability p that a test is positive is the same for all people and that these events are independent.
Find the probability that the test for a pooled sample of k people will be positive.
What is the expected value of the number X of tests necessary under plan (2)? (Assume that N is divisible by k.)
For small p, show that the value of k which will minimize the expected number of tests under the second plan is approximately \(1/\sqrt{p}\)
\[ P[ \text{ k people are all negative}] = (1 - p)^k\] The pooled sample will test positive if not all \(k\) people are negative. Equivalently if at least one person tests positive. This event is the complement of the event that all people in the sample are negative.
The answer is the probability of a positive test is:
\[ P[ \text{ pooled sample of k people is positive }] = 1 - (1 - p)^k \]
\[ X = \frac{N}{k} [ 1 + (1 - ( 1 - p )^k )k ]\]
\[ X = N [ \frac{1}{k} + 1 - (1-p)^k]\] When \(p\) is small, we can use the Taylor approximation of \[(1-p)^k \sim 1 - kp\]
This gives the identity:
\[ X = N [ \frac{1}{k} + 1 - ( 1 -kp)]\] \[X = N [ \frac{1}{k}+ kp ]\]
Differentiating with respect to \(k\) and setting the first derivative to zero allows us to find the minima:
\[\frac{ \partial X}{\partial k} = N[ -\frac{1}{k^2} + p] = 0 \implies k^2=\frac{1}{p} \implies k = \frac{1}{\sqrt{p}}\]