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

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,

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

P[X=0] = 0.0115

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

P[X=1] = 0.0576

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

P[X=2] = 0.1369

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

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