Question

Is the package late? A shipping company claims that 90% of its shipments

arrive on time. Suppose this claim is true. If we take a random sample of 20 shipments made by the company, what’s the probability that at least 1 of them arrives late?

Short answer

The probability that at least one shipment arrives late is$0.8784$

Step-by-step solution

Given information

$90\%$shipments arrive on time.

$20$ shipments are selected at random.

Calculation

Two events are independent if the probability of one event's occurrence has no bearing on the probability of the other event's occurrence.

For independent events, the multiplication rule is as follows:

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}\cap \mathrm{B})=P(\mathrm{A})\times P(\mathrm{B})$

The complement rule states that

$P\left({\mathrm{A}}^{\mathrm{c}}\right)=P\left(\text{not}\mathrm{A}\right)=1-P(\mathrm{A})$

Let

$A$: One shipment arrives on time

$B$: At least 1 of the 20 shipments arrives late

$A^c:$One shipment arrives late

$B^c$: None of the 20 shipments arrives late

Now,

Probability for shipment arriving on time,
$P(\mathrm{A})=90\%=0.90$

Because the 20 shipments are chosen at random, it is more convenient to assume that they are unrelated to one another.

Apply the multiplication rule for independent events to the probability that none of the 20 shipments arrive late:

$P\left({\mathrm{B}}^{\mathrm{c}}\right)=\underset{\text{20repetitions}}{\underbrace{P(\mathrm{A})\times P(\mathrm{A})\times \dots \times P(\mathrm{A})}}=\left(P\right(\mathrm{A}){)}^{20}=(0.90{)}^{20}\approx 0.1216$

Apply the complement rule:

$P\left(B\right)=P\left({\left(B^{\mathrm{c}}\right)}^{\mathrm{c}}\right)=1-P\left({\mathrm{B}}^{\mathrm{c}}\right)=1-0.1216=0.8784$

Therefore, From random samples of 20 shipments, the probability that at least 1 shipment arrives late is $0.8784.$