Respuesta :
Answer:
(a) P (X ≥ 4) = 0.972
(b) E (X) = 20
Step-by-step 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:
[tex]f(x)={x-1\choose r-1}p^{r}(1-p)^{x-r}[/tex]
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)
[tex]=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[/tex]
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:
[tex]E(X)=\frac{r}{p}[/tex]
The expected number of people to be tested before two with gene are detected is:
[tex]E(X)=\frac{r}{p}=\frac{2}{0.1}=20[/tex]
Thus, the expected number of people to be tested before two with gene are detected is 20.
