Question

A company owns 400 laptops. Each laptop has an 8% probability of not working. You randomly select 20 laptops for your salespeople. 1.What is the likelihood at most 6 will be broken? 2. What is the likelihood that less than 18 will be broken? 3. What is the likelihood that more than 3 will be broken?

Answer 1

we have probability of broken laptop = 8% or 8/100 = 0.08

sample size(n) = 20

(1) probability that at most 6 will be broken = binomcdf(n,p, largest x value)

here we have n=20, p=0.08 and largest x value is 6

setting the value, we get

probability that at most 6 will be broken = binomcdf(20,0.08,6) = 0.9994(rounded to 4 decimal places)

(2) Probability that less than 18 will be broken = probability that at most 17 will be broken

so, we have n = 20, p = 0.08 and largest x value = 17

thus, we get

Probability that less than 18 will be broken = probability that at most 17 will be broken

= binomcdf(20,0.08,17) = 0.9999 (rounded to 4 decimal places)

(3) Probability of more than 3 broken laptop = 1 - P(at most 3 broken)

and we can write, P(at most 3 broken) = binomcdf(n,p, largest x value) = binomcdf(20,0.08,3)= 0.9294

So, the probability of more than 3 broken laptop = 1 - P(at most 3 broken) = 1 - 0.9294 = 0.0706 (rounded to 4 decimal places)