Respuesta :
Answer:
a) There is a 12.11% probability that exactly 1 man has the marker.
b) There is a 85.07% probability that more than 1 has the marker.
Step-by-step explanation:
There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.
Binomial probability
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And [tex]\pi[/tex] is the probability of X happening.
In this problem, we have that:
30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so [tex]\pi = 0.30[/tex]
(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?
10 men, so [tex]n = 10[/tex]
We want to find P(X = 1). So:
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211[/tex]
There is a 12.11% probability that exactly 1 man has the marker.
(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?
That is [tex]P(X > 1)[/tex]
We have that:
[tex]P(X \leq 1) + P(X > 1) = 1[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]
We also have that:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282[/tex]
So
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493[/tex]
Finally
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507[/tex]
There is a 85.07% probability that more than 1 has the marker.
