Answer:
a. The probability is 0.735
b. The probability is 0.6296
c. The probability is 0
Step-by-step explanation:
If we assume a binomial process, the probability that x customer are willing to switch companies is:
![P(x)=nCx*p^{x}*(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3DnCx%2Ap%5E%7Bx%7D%2A%281-p%29%5E%7Bn-x%7D)
nCx is calculated as:
![nCx=\frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=nCx%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Where n are the 20 cable customers and p is the probability 0.2 that the cable customer are willing to switch companies.
Then, P(x) is:
![P(x)=20Cx*0.2^{x}*(1-0.2)^{20-x}](https://tex.z-dn.net/?f=P%28x%29%3D20Cx%2A0.2%5E%7Bx%7D%2A%281-0.2%29%5E%7B20-x%7D)
The probability that between 2 and 5 (inclusive) customers are willing to switch companies is:
P(2≤x≤5) = P(2) + P(3) + P(4) + P(5)
Where P(2), P(3), P(4) and P(5) are equal to:
![P(2)=20C2*0.2^{2}*(1-0.2)^{20-2}=0.1369](https://tex.z-dn.net/?f=P%282%29%3D20C2%2A0.2%5E%7B2%7D%2A%281-0.2%29%5E%7B20-2%7D%3D0.1369)
![P(3)=20C3*0.2^{3}*(1-0.2)^{20-3}=0.2054](https://tex.z-dn.net/?f=P%283%29%3D20C3%2A0.2%5E%7B3%7D%2A%281-0.2%29%5E%7B20-3%7D%3D0.2054)
![P(4)=20C4*0.2^{4}*(1-0.2)^{20-4}=0.2182](https://tex.z-dn.net/?f=P%284%29%3D20C4%2A0.2%5E%7B4%7D%2A%281-0.2%29%5E%7B20-4%7D%3D0.2182)
![P(5)=20C5*0.2^{5}*(1-0.2)^{20-5}=0.1745](https://tex.z-dn.net/?f=P%285%29%3D20C5%2A0.2%5E%7B5%7D%2A%281-0.2%29%5E%7B20-5%7D%3D0.1745)
So, P(2≤x≤5) is:
P(2≤x≤5) = 0.1369 + 0.2054 + 0.2182 + 0.1745 = 0.735
At the same way, the probability that less than 5 customers are willing to switch is:
P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)
P(x<5)=0.6296
Finally, the probability that more than 16 customers are willing to switch is:
P(x>16)=P(17)+P(18)+P(19)+P(20)
P(x>16)=0