Chapter 5.3, Problem 91E

To determine

Probability for at least one of shipment from random sample of 20 shipments arrives late.

Answer to Problem 91E

Probability that at least one shipment arrives late is 0.8784.

Explanation of Solution

Given information:

90% shipments arrive on time.

20 shipments are selected at random.

Calculations:

Two events are independent, if the probability of occurrence of one event does not affect the probability of occurrence of other event.

According to multiplication rule for independent events,

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

According to complement rule,

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

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,

  $P\left(\text{A}\right)=90\%=0.90$

Since the 20 shipments are selected at random, it would be more convenient to assume that shipments are independent of each other.

Thus,

For probability that none of 20 shipments arrives late, apply multiplication rule for independent events:

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

Apply the complement rule:

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

Thus,

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