Question

A computer retail store has 1414 personal computers in stock. A buyer wants to purchase 33 of them. Unknown to either the retail store or the buyer, 33 of the computers in stock have defective hard drives. Assume that the computers are selected at random.
A) In how many different ways can the 3 computers be chosen? 120
B) What is the probability that exactly one of thecomputers will be defective?
C) What is the probability that at least one of thecomputers selected is defective?

Answer 1

Answer:

a) 364 ways

b) 45.33% probability that exactly one of the computers will be defective.

c) 54.67% probability that at least one of the computers selected is defective.

Step-by-step explanation:

The computers are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

14 computers, so N = 14.

3 defective, so k = 3.

3 will be purchesed, so n = 3.

A) In how many different ways can the 3 computers be chosen?

3 from a set of 14. So

$C_{14,3} = \frac{14!}{3!(14-3)!} = 364$

364 ways

B) What is the probability that exactly one of thecomputers will be defective?

This is P(X = 1).

$P(X = 1) = h(1,14,3,3) = \frac{C_{3,1}*C_{11,2}}{C_{14,3}} = 0.4533$

45.33% probability that exactly one of the computers will be defective.

C) What is the probability that at least one of the computers selected is defective?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. Then

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

In which

$P(X = 0) = h(0,14,3,3) = \frac{C_{3,0}*C_{11,3}}{C_{14,3}} = 0.4533$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4533 = 0.5467$

54.67% probability that at least one of the computers selected is defective.