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$81\%$of the incoming calls are for medical help.

$4$incoming calls to the station are chosen at random.

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,

$P(AandB)=P(A鈭�B)=P\left(A\right)脳P\left(B\right)$

Let

A: One incoming call is for medical help

B: $4$incoming calls are for medical help

Now,

Probability for the incoming call is for medical help,

$P\left(A\right)=81\%=0.81$

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

Thus,

For probability that $4$incoming calls are for medical help, apply multiplication rule for independent events:

role="math" localid="1664275493508" $$\begin{gathered} P\left(B\right)=P\left(A\right)脳P\left(A\right)脳P\left(A\right)脳P\left(A\right) \\ =(P{\left(A\right))}^4 \\ ={(0.81)}^4 \\ 鈮�0.4305 \end{gathered}$$

Thus,

Probability for the randomly selected all $4$incoming calls are for medical help is approx.$0.4305$

According to complement rule,

$P\left(A^c\right)=P(notA)=1-P\left(A\right)$

Let

$B$4 incoming calls are for medical help

$B^c:$None of the $4$incoming calls are for medical help

From Part (a),

We have

Probability for randomly selected all 4 incoming calls are for medical help,

$P\left(B\right)鈮�0.4305$

We have of find the probability for at least $1$of the calls is not for medical help.

That means

None of the $4$incoming calls is for medical help.

Apply the complement rule:

$$\begin{gathered} P\left(B^c\right)=1-P\left(B\right) \\ =1-0.4305 \\ =0.5695 \end{gathered}$$

Thus,

Probability that at least $1$of the calls is not for medical help is$0.5695$

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,

$P(AandB)=P(A鈭�B)=P\left(A\right)脳P\left(B\right)$

In Part (a),

Multiplication rule for independent events has been used.

When $4$consecutive calls are chosen, there may be some possibility that occurrence of these $4$calls would be same.

That means

$4$consecutive calls may be for the same accident when compared to the randomly chosen $4$calls.

Thus,

This will affect the probability for medical call if these $4$calls are about the same occurrence.

This implies

The incoming calls will be no longer independent.

Thus,

Use of the multiplication for independent events would be inappropriate.