Question

The Alvin Secretarial Service procures temporary office personnel for major corporations. They have found that 60% of their invoices are paid within ten working days. A random sample of 18 invoices is checked. What is the probability that at least 6 of the invoices will be paid within ten working days? Round your answer to four decimal places.

Answer 1

Answer:

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days

Step-by-step explanation:

For each invoice, there are only two possible outcomes. Either they are paid within 10 working days, or they are not. The probability of an invoice being paid within 10 working days is independent of other invoices. 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.

60% of their invoices are paid within ten working days.

This means that $p = 0.6$

A random sample of 18 invoices is checked.

This means that $n = 18$

What is the probability that at least 6 of the invoices will be paid within ten working days?

Either less than 6 are paid within 10 working days, or at least 6 are paid. The sum of the probabilities of these events is 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$. So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Then

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

$P(X = 0) = C_{18,0}.(0.6)^{0}.(0.4)^{18} \approx 0$

$P(X = 1) = C_{18,1}.(0.6)^{1}.(0.4)^{17} \approx 0$

$P(X = 2) = C_{18,2}.(0.6)^{2}.(0.4)^{16} \approx 0$

$P(X = 3) = C_{18,3}.(0.6)^{3}.(0.4)^{15} = 0.0002$

$P(X = 4) = C_{18,4}.(0.6)^{4}.(0.4)^{14} = 0.0011$

$P(X = 5) = C_{18,5}.(0.6)^{5}.(0.4)^{13} = 0.0045$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0 + 0 + 0 + 0.0002 + 0.0011 + 0.0045 = 0.0058$

Finally

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.0058 = 0.9942$

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days