Question

Historical data indicates that only 20% of cable customers are willing to switch companies. If a binomial process is assumed, then in a sample of 20 cable customers, what is the probability that no more than 3 customers would be willing to switch their cable?

a. 0.15
b. 0.411
c. 0.85
d. 0.20

Answer 1

Answer:
`P[X\leq 3]` = =0.4114

Step-by-step explanation:

Given that ;

p=0.20

n = 20

Therefore we can compute the probability as;

`P[X\leq 3]=P[X=0]+P[X=1]+P[X=2]+P[X=3]`

where,

`P[X=0]=\binom{20}{0}*(0.20)^(0)*(1-0.20)^(20)`

P[X=0] = 0.0115

`P[X=1] = \binom{20}{1}*(0.20)^(1)*(1-0.20)^(19)`

P[X=1] = 0.0576

`P[X=2] = \binom{20}{2}*(0.20)^(2)*(1-0.20)^(18)`

P[X=2] = 0.1369

`P[X=3] = \binom{20}{3}*(0.20)^(3)*(1-0.20)^(17)`

P[X=3] = 0.2054

Therefore;

`P[X\leq 3]=P[X=0]+P[X=1]+P[X=2]+P[X=3]`

=0.0115+0.0576 +0.1369 + 0.2054

=0.4114