Question

A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them. Suppose 9 employees are selected randomly and will be interviewed about company loyalty.
A. What is the probability that none of the 9 employees will say their company is loyal to them?

c. What is the probability that 4 of the 9 employees will say their company is loyal to them?

Answer 1

Answer:

(A) 0.999996

(B) 0.11680

Step-by-step explanation:

We are given that a Randstad Harris interactive survey reported that 25% of employees said their company is loyal to them.

And 9 employees are selected randomly and interviewed about company loyalty.

The Binomial probability distribution is given by;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....$

where, n = number of trials (samples) taken

             r = number of success

             p = probability of success

In our question; n = 9 , p = 0.25 (as employees saying their company is loyal to them is success to us)

(A) Probability that none of the 9 employees will say their company is loyal to them = 1 - Probability that all 9 employees will say their company is loyal to them

= 1 - P(X = 9)  { As here number of success is 9 }

= 1 - $\binom{9}{9}0.25^{9}(1-0.25)^{9-9}$ = 1 - $0.25^{9}$ = 0.999996

(B) Probability that 4 of the 9 employees will say their company is loyal to them = P(X = 4)

    P(X = 4) = $\binom{9}{4}0.25^{4}(1-0.25)^{9-4}$

                  = $126*0.25^{4}*0.75^{5}$ = 0.11680