Answer:
a)There is a 4.88% probability that none is concerned that employers are monitoring phone calls.
b)There is a 7.89% probability that all are concerned that employers are monitoring phone calls.
c)There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.
Step-by-step explanation:
The binomial probability is the probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment).
It is given by the following formula:
![P = C_{n,x}.p^{n}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bn%7D.%281-p%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 p is the probability of a success.
In this problem, a success is being concerned that employers are monitoring phone calls.
53% of adults are concerned that employers are monitoring phone calls, so ![p = 0.53](https://tex.z-dn.net/?f=p%20%3D%200.53)
(a) Out of four adults, none is concerned that employers are monitoring phone calls.
Four adults, so
.
Is the probability of 0 successes, so x = 0.
![P = C_{n,x}.p^{n}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bn%7D.%281-p%29%5E%7Bn-x%7D)
![P = C_{4,0}.(0.53)^{0}.(0.47)^{4}](https://tex.z-dn.net/?f=P%20%3D%20C_%7B4%2C0%7D.%280.53%29%5E%7B0%7D.%280.47%29%5E%7B4%7D)
![P = 0.0488](https://tex.z-dn.net/?f=P%20%3D%200.0488)
There is a 4.88% probability that none is concerned that employers are monitoring phone calls.
(b) Out of four adults, all are concerned that employers are monitoring phone calls.
Four adults, so
.
Is the probability of 4 successes, so x = 4.
![P = C_{n,x}.p^{n}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bn%7D.%281-p%29%5E%7Bn-x%7D)
![P = C_{4,0}.(0.53)^{4}.(0.47)^{0}](https://tex.z-dn.net/?f=P%20%3D%20C_%7B4%2C0%7D.%280.53%29%5E%7B4%7D.%280.47%29%5E%7B0%7D)
![P = 0.0789](https://tex.z-dn.net/?f=P%20%3D%200.0789)
There is a 7.89% probability that all are concerned that employers are monitoring phone calls.
(c) Out of four adults, exactly two are concerned that employers are monitoring phone calls.
Four adults, so
.
Is the probability of 4 successes, so x = 2.
![P = C_{n,x}.p^{n}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bn%7D.%281-p%29%5E%7Bn-x%7D)
![P = C_{4,2}.(0.53)^{2}.(0.47)^{2}](https://tex.z-dn.net/?f=P%20%3D%20C_%7B4%2C2%7D.%280.53%29%5E%7B2%7D.%280.47%29%5E%7B2%7D)
![P = 0.3723](https://tex.z-dn.net/?f=P%20%3D%200.3723)
There is a 37.23% probability that exactly two are concerned that employers are monitoring phone calls.