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$90\%$shipments arrive on time.

$20$ shipments are selected at random.

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:

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}鈭�\mathrm{B})=P(\mathrm{A})脳P(\mathrm{B})$

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:

$P\left({\mathrm{B}}^{\mathrm{c}}\right)=\underset{\text{20repetitions}}{\underset{铃�}{P(\mathrm{A})脳P(\mathrm{A})脳鈥�脳P(\mathrm{A})}}=\left(P\right(\mathrm{A}){)}^{20}=(0.90{)}^{20}鈮�0.1216$

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 $0.8784.$