Question

Most internet service provider (ISPs) attempt to provide a large enough service so that customers seldom encounter a busy signal. Suppose that the customers of one ISP encounter busy signals 4% of the time. During the week a customer of this ISP called 25 times. What is the probability that she encountered at lease one busy signal?

This is from a midterm and i answered 0.3754 and got it wrong, if you could answer and show work it would be greatly appreciated.  

Answer 1

Solution

Given that ,

p = 4% = 0.04

1 - p = 1 - 0.04 = 0.96

n = 25

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X≥1)= 1 - P(X<1)

= 1 - P(X=0)

P(X≥1) = 1 - (((25! / 0! (25-0)!) * 0.04⁰ * (0.96)^25-0 )

= 1 - ( ((25! / 0! (25)!) * 0.04⁰ * (0.75)²⁵ )

= 1 - 0.3604

= 0.6396

Probability = 0.6396