Question

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.
(round to 4 decimal places. leave your answer in decimal form.)
(b) whats the probability that at least 1 of the calls is not for medical help?
(round to 4 decimal places. leave your answer in decimal form.)
(c) explain why the calculation in part (a) may not be valid if we choose 4 consecutive calls to the station.
the calculation in part (a) might not be valid because the 4 consecutive calls being medical may not be events.

Explanation:

Step1: Calculate probability for part (a)

Let \(p = 0.81\) (probability of a call being for medical help). The probability that all \(n = 4\) calls are for medical help is \(P(X = 4)\) using the binomial probability formula \(P(X=k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here \(n = 4,k = 4\), so \(C(4,4)=\frac{4!}{4!(4 - 4)!}=1\). Then \(P(X = 4)=1\times(0.81)^{4}\times(1 - 0.81)^{0}=(0.81)^{4}\approx0.4305\)

Step2: Calculate probability for part (b)

The probability that at least 1 call is not for medical help is \(P(X\geq1)=1 - P(X = 0)\). First, find \(P(X = 0)\) with \(n = 4,k = 0\), \(C(4,0)=\frac{4!}{0!(4-0)!}=1\), \(p = 0.81\), \(1-p=0.19\). So \(P(X = 0)=1\times(0.81)^{0}\times(0.19)^{4}=(0.19)^{4}\approx0.0013\). Then \(P(X\geq1)=1 - 0.0013 = 0.9987\)

Answer:

(a) \(0.4305\)
(b) \(0.9987\)