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.
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And
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 ![\pi = 0.30](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.30)
(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 ![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
We want to find P(X = 1). So:
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
![P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B10%2C1%7D.%280.30%29%5E%7B1%7D.%280.7%29%5E%7B9%7D%20%3D%200.1211)
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 ![P(X > 1)](https://tex.z-dn.net/?f=P%28X%20%3E%201%29)
We have that:
![P(X \leq 1) + P(X > 1) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%201%29%20%2B%20P%28X%20%3E%201%29%20%3D%201)
![P(X > 1) = 1 - P(X \leq 1)](https://tex.z-dn.net/?f=P%28X%20%3E%201%29%20%3D%201%20-%20P%28X%20%5Cleq%201%29)
We also have that:
![P(X \leq 1) = P(X = 0) + P(X = 1)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%201%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29)
![P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B10%2C0%7D.%280.30%29%5E%7B0%7D.%280.7%29%5E%7B10%7D%20%3D%200.0282)
So
![P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493](https://tex.z-dn.net/?f=P%28X%20%5Cleq%201%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%3D%200.0282%20%2B%200.1211%20%3D%200.1493)
Finally
![P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507](https://tex.z-dn.net/?f=P%28X%20%3E%201%29%20%3D%201%20-%20P%28X%20%5Cleq%201%29%20%3D%201%20-%200.1493%20%3D%200.8507)
There is a 85.07% probability that more than 1 has the marker.