Question

A company finds that one out of every 5 workers it hires turns out to be unsatisfactory. Assume that the satisfactory performance of any hired worker is independent of that of any other hired workers. If the company hires 15 people, what is the probability that the following number of people will turn out to be satisfactory? (Round your answer to six decimal places.)

exactly 9

Answer 1

Solution:

From the given information,

probability of unsatisfactory = 1/5 = 0.2

probability of satisfactory = 1 - 0.2 = 0.8

n = 15 (No. of trials)

Let X be the number of  satisfactory people int his sample.

X follows Binomial( 15 , 0.8)

Find P(Exactly 9 people will turn out to be satisfactory)

i.e. Find P(X = 9)

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x}  

P(X = 9) = (₁₅ C ₉) * 0.8⁹ * (1 - 0.8)^15-9

= (15!/9!*6!) * 0.8⁹ * 0.2⁶

=  0.042993