# Probability for testing positive without having disease

This is a question I came across recently, any help would be greatly appreciated!:

A biostatistician is testing genes to determine their association to a certain disease. Each test for a given gene has a $5$% chance of a false positive for association (that is, about $5$% of tests performed on genes with no association will have an outcome showing association).

The biostatistician performs the test on $4$ separate genes so that the outcome of each test is independent (this means that the outcome of different tests don’t affect one another).

$(a)$ Define a suitable outcome space for this experiment.

$(b)$ Define an appropriate probability function for the outcome space of Part $(a)$ to answer the following question: If in reality, none of the 4 tested genes have an association with the disease, what is the chance that at least one of the tests will (falsely) show an association?

I am not sure about several things:

$1.$ Would the outcome space be $(+,-)$ for the test (testing pos. and neg.) or would it be all the combinations for the $4$ subjects testing positive and negative? (I am leaning towards the second one but would just like to make sure)

$2$. I understand that we have been given that $P(+|D') = 0.05$ and it is my understanding that we are being asked to find $P(X \geq 1)$ where $X$ is the number of false positives but I am not sure how to go about this at all.

Assigning $P(D) = 0.01$ and simply calculating $4 * 0.01 * 0.05$ seems too simple and wrong.

I'm really struggling with probability concepts so I would be grateful for any explanations!

Stripped of its context, I believe your question is the following: you are conducting a series of independent trials. The probability that a single trial succeeds is $p=.05$. What is the probability that you get at least one success out of four trials?
Assuming this is correct then your sample space has $16=2^4$ elements (as each trial has two possible outcomes).
To compute the probability you want, it is easier to compute the probability of the complementary event (namely "all four trials are failures"). The probability that a single trial is a failure is $1-p=.95$ so the probability that all four are failures is $.95^4\sim .8145$. It follows that the probability that at least one trial is a success is $$1-.95^4\sim .1855$$