A consumer organization estimates that over a 1-year period 15% of cars will nee

Question

A consumer organization estimates that over a 1-year period 15% of cars will need to be repaired once. 8% will need repairs twice, and 3% will require three or more repairs What is the probability that a car chosen at random will need a) no repairs? b) no more than one repair? C) some repairs? a) The probability that a car will require no repairs is .(Do not round.) b) The probability that a car will require no more than one repair is (Do not round.) c) The probability that a car will require some repairs is (Do not round.) Enter your answer in each of the answer boxes.

Explanation / Answer

one repair = 0.15

two repairs = 0.08

three or more repairs = 0.03

Total = 0.26

So, no repair = 1 - 0.26 = 0.74

(a) The probability that a car will require no repair = 0.74

(b) The probability that a car will require no more than one repair = P(one repair) + P(no repair) = 0.15 + 0.74 = 0.89

(c) The probability that a car will require some repair = P(one repair) + P(two repairs) + P(3 or more repairs) = 0.15 + 0.08 + 0.03 = 0.26

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