Question

QUESTION 7 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. What is the probability that none of the 9 employees will say their company is loyal to them? g

Answer 1

Answer:

7.51% probability that none of the 9 employees will say their company is loyal to them.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of employees said their company is loyal to them.

This means that $p = 0.25$

9 employees are selected randomly

This means that $n = 9$

What is the probability that none of the 9 employees will say their company is loyal to them?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751$

7.51% probability that none of the 9 employees will say their company is loyal to them.