Question

In clinIcal study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability that a person carries the gene is 0.1.

a) what is the probability four or more people will have to be tested before two with the gene are detected?b) How many people are expected to be tested before two with gene are detected?

Answer 1

(a) P (X ≥ 4) = 0.972

(b) E (X) = 20

Explanation:

Let X = number of people tested to detect the presence of gene in 2.

Then the random variable X follows a Negative binomial distribution with parameters r (number of success) and p probability of success.

The probability distribution function of X is:

`f(x)={x-1\choose r-1}p^(r)(1-p)^(x-r)`

Given: r = 2 and p = 0.1

(a)

Compute the probability that four or more people will have to be tested before two with the gene are detected as follows:

P (X ≥ 4) = 1 - P (X = 3) - P (X = 2)

`=1-[{3-1\choose 2-1}(0.1)^(2)(1-0.1)^(3-2)]-[{2-1\choose 2-1}(0.1)^(2)(1-0.1)^(2-2)]\\=1-0.018-0.01\\=0.972`

Thus, the probability that four or more people will have to be tested before two with the gene are detected is 0.972.

(b)

The expected value of a negative binomial random variable X is:

`E(X)=(r)/(p)`

The expected number of people to be tested before two with gene are detected is:

`E(X)=(r)/(p)=(2)/(0.1)=20`

Thus, the expected number of people to be tested before two with gene are detected is 20.