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.)

Answer 1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.85

N = number of trials = 3

R = desired success

A)

Here P = 1 - 0.85 = 0.15 (as we are talking about emergencies)

P(3) = 3c3*(0.15^3)*(1-0.15)^3-3

= 0.003375

B)

P(2 or more) = p(2) + p(3)

P = 0.85

= 0.93925

C)

P(at least 1) = 1 - p(0)

0.96 = 1 - nc0*(0.15^0)*(1-0.15)^n-0

0.85^n = 0.04

Taking log on both sides

nlog0.85 = log0.04 (as loga^n = nloga)

n = log0.04/log0.85

n = 20