Question

Fire or medical? Many fire stations handle more emergency calls for medical help than for fires. At one fire station, 81% of incoming calls are for medical help. Suppose we choose 4 incoming calls to the station at random.

a. Find the probability that all 4 calls are for medical help.

b. What’s the probability that at least 1 of the calls is not for medical help?

c. Explain why the calculation in part (a) may not be valid if we choose 4 consecutive calls to the station.

Short answer

Part a) Probability for the randomly chosen 4 calls for medical help is approx.$0.4305.$

Part b) Probability that at least 1 of the calls is not for medical help is$0.5695.$

Part c) It is not necessary that calls are independent of each other.

Step-by-step solution

Part a) Step 1: Given information

Medical assistance is requested in $81\%$of incoming calls.

At random, $4$ incoming calls to the station are chosen.

Part a) Step 2: 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:

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

$A:$One incoming call is for medical help

$B:$incoming calls are for medical help

Now, the likelihood of an incoming call for medical assistance is high.

$P(\mathrm{A})=81\%=0.81$

Because incoming calls are chosen at random, it would be more convenient to assume that they are unrelated to one another.

Apply the multiplication rule for independent events to the probability that $4$incoming calls are for medical assistance:

$P(\mathrm{B})=P(\mathrm{A})\times P(\mathrm{A})\times P(\mathrm{A})\times P(\mathrm{A})=\left(P\right(\mathrm{A}){)}^4=(0.81{)}^4\approx 0.4305$

Therefore, the probability for the randomly selected $4$incoming calls for medical help is approx. $0.4305.$

Part b) Step 1: Given information

Medical assistance is requested in $81\%$of incoming calls.

At random, $4$incoming calls to the station are chosen.

Part b) Step 2: Calculation

According to the complement rule,

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

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 a probability that all four incoming calls are for medical assistance.

$P\left(B\right)\approx 0.4305$

We have a probability that all four incoming calls are for medical assistance.

This means that none of the four incoming calls are for medical assistance.

Use the complement rule to help you:

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

Therefore, the probability that at least 1 of the calls is not for medical help is $0.5695.$

Part c) Step 1: Given information

Medical assistance is requested in $81\%$of incoming calls.

At random, $4$incoming calls to the station are chosen.

Part c) Step 2: 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:

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

In part (a)

For independent events, the multiplication rule was used.

When $4$consecutive calls are chosen, there is a chance that the occurrence of these four calls will be the same.

That means,

When compared to the randomly chosen $4$calls, $4$consecutive calls could be for the same accident.

Thus,

If these four calls are about the same occurrence, this will have an impact on the likelihood of a medical call.

Therefore, It would be inappropriate to use multiplication for independent events.