I am having issues with understanding a question on an assignment:

A test for the presence of a certain disease has probability 0.05 of giving a false-positive reading and probability 0.04 of giving a false-negative result. Suppose that four individuals are tested, three of whom do not have the disease and one of whom does. Let X= the number of positive readings.

  1. Does X have a binomial distribution?
  2. Find the probability that only one of the four test results is positive.

Edit: I'm not sure if this is correct, but I have the probability setup like this: P(A->P)P(B->N)P(C->N)P(D->N)+P(A->N)P(B->P)P(C->N)P(D->N)+P(A->N)P(B->N)P(C->P)P(D->N)+P(A->N)P(B->N)P(C->N)P(D->P)

Where A,B and C do not have the disease. D has the disease. (A->P means A tests positive)


2 Answers 2


What you are looking for is the probability generating function. This functions allows you to plug in the values given in your exercise and calculate the probabilities needed. A nice derivation of the probability generating function of the binomial distribution can be found under


If you are interested it will additionally provide you with the expected value and variance estimation.

  1. A binomial distribution models the number of positive outcomes in a number of independent true or false events (Bernoulli events) that each have the same probability of being true. You should ask yourself: for the four individuals being tested, does each person have the same probability of testing positive?

  2. If you have two independent events, the probability of any two events occurring is the probability of those two events occurring individually multiplied by each other, then multiplied by the number of permutations that it can occur in. Apply this to the case of four events. You should take note that this is exactly the principle that the binomial distribution is built off of. So if you only had two people, person A and person B, then $$\begin{align*} \mathbb{P}[&\text{exactly one person tests positive}] \\ &= \mathbb{P}[A \text{ tests positive}]\mathbb{P}[B \text{ tests negative}] + \mathbb{P}[A \text{ tests negative}]\mathbb{P}[B \text{ tests positive}] \end{align*}$$

  • $\begingroup$ Thanks for the reply, I'm thinking that it is not a binomial distribution since 3 have the disease and 1 doesn't. $\endgroup$
    – Anom316
    Commented Sep 30, 2014 at 2:55

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