Question

A marketing firm is considering making up to three new hires. Given its specific needs, the management feels that there is a 60% chance of hiring at least two candidates. There is only a 6% chance that it will not make any hires and a 16% chance that it will make all three hires.
Required:
a. What is the probability that the firm will make at least one hire?
b. Find the expected value and the standard deviation of the number of hires.

Answer 1

Answer:

a. 0.94 = 94% probability that the firm will make at least one hire.

b. The expected value of the number of hires is 1.7, and the standard deviation is 0.8062.

Step-by-step explanation:

There is only a 6% chance that it will not make any hires

This means that P(X = 0) = 0.06.

16% chance that it will make all three hires.

This means that $P(X = 3) = 0.16$

60% chance of hiring at least two candidates.

This means that:

$P(X = 2) + P(X = 3) = 0.6$

$P(X = 2) + 0.16 = 0.6$

$P(X = 2) = 0.44$

Probability of one hire:

The sum of all probabilities, from no hires to three hires, is 1. So

$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1$

$0.06 + P(X = 1) + 0.16 + 0.44 = 1$

$P(X = 1) = 0.66 = 1$

$P(X = 1) = 0.34$

a. What is the probability that the firm will make at least one hire?

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.06 = 0.94$

0.94 = 94% probability that the firm will make at least one hire.

b. Find the expected value and the standard deviation of the number of hires.

Expected value:

$E(X) = 0.06*0 + 0.34*1 + 0.44*2 + 0.16*3 = 1.7$

Standard deviation:

$S(X) = \sqrt{0.06*(0-1.7)^2 + 0.34*(1-1.7)^2 + 0.44*(2-1.7)^2 + 0.16*(3-1.7)^2} = 0.8062$

The expected value of the number of hires is 1.7, and the standard deviation is 0.8062.