Question

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 three calls.

(a) What is the probability that all three calls are, in fact, emergencies? (Round your answer to three decimal places.)


(b) What is the probability that two or more calls are not emergencies? (Round your answer to three decimal places.)


(c) 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? (Enter your answer as a whole number.)
calls

Answer 1

Solution:-

p = 0.85

a) The probability that all three calls are, in fact, emergencies is 0.6141.

n = 3, x = 3

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x = 3) = 0.6141

(b) The probability that two or more calls are not emergencies is

n = 3, x = 2

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 2) = 0.9393

(c) The number of calls we need to answer on 911 operators to be 96% (or more) sure that at least one call is, in fact, an emergency is 20.

p₁ = 1 - 0.85 = 0.15

x = 1

P(x > 1) = 0.96

By using binomial

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

n = 20

answered by: anonymous