Solved-20. Public Safety: 911 Calls The Denver Post reported that a recent audit of Los Angeles 911 calls showed that 85 were

ByDr. Raju Chaudhari

Oct 6, 2020

20.Public Safety: 911 Calls The Denver Post reported that a recent audit of Los Angeles 911 calls showed that 85% were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls.

(a) What is the probability that all four calls are, in fact, emergencies?
(b) What is the probability that three or more calls are not emergencies?
(c) Quota Problem How many calls n would the 911 operators need to answer to be 96% (or more) sure that at least one call is, in fact, an emergency?

Solution

Let $X$ denote the number of emergency calls out of 4 calls.

Let $p$ denote the probability of emergency call. Given that the probability that the call is not emergency is 0.85. Thus $p= 1- 0.85 = 0.15$.

$X\sim B(n=4, p=0.15)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{4}{x} (0.15)^x (1-0.15)^{4-x},\\ & \qquad x=0,1,\cdots, 4. \end{aligned}

a) The probability that all four calls are, in fact, emergencies

 \begin{aligned} P(X= 4) & =\binom{4}{4} (0.15)^{4} (1-0.15)^{4-4}\\ & = 0.0005\\ \end{aligned}

b) Three or more calls are not emergencies means one or none are emergency.

Thus the probability that three or more calls are not emergencies is equal to probability that one or none beging emergency.

 \begin{aligned} P(X\leq 1) & =\binom{4}{0} (0.15)^{0} (1-0.15)^{4-0}+\binom{4}{1} (0.15)^{1} (1-0.15)^{4-1}\\ & = 0.8905\\ \end{aligned}

c) The probability of $n$ calls received being all non-emergencies = $(0.85)^n$

For a 0.96 probability that at least one is an emergency, you want the probability that they are all non-emergencies to be lower than 0.04.

That is,

 \begin{aligned} & (0.85)^n < 0.04 \\ &\Rightarrow n \log (0.85) < \log (0.04) \\ &\Rightarrow n < \log (0.04) / \log (0.85)\\ & \Rightarrow n < 19.8 \end{aligned}

So n must exceed 19.8, i.e. n must be atleast 20.