# Help for Binomial Distribution question

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. 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*}