The Denver Post reported that a recent Audit of Los Angeles 911 calls showed tha

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 two calls. (a) What is the probability that all two calls are, in fact, emergencies? (Round your answer to three decimal places.)(b) What is the probability that one or more calls are not emergencies?(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? and show how you got it please!

Explanation / Answer

Here,

P(emergency) = 1 - 0.85 = 0.15.

a)

P(both are emergency) = 0.15*0.15 = 0.0225 [ANSWER]

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b)

P(at least one is not an emergency) = 1 - P(both are emergencies)

= 0.9775 [ANSWER]
******************

c)

P(at least one emergency) = 1 - P(none is an emergency)

Let n = the number of calls. Thus,

P(at least one emergency) = 1 - 0.86^n = 0.96

1 - 0.86^n = 0.96

0.86^n = 0.04

n = ln (0.04) / ln (0.86) = 21.34209092

rounding up,

n = 22 [ANSWER]

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