A consumer organization estimates that over a 1-year period 1919% of cars will n

ID: 3052200 • Letter: A

Question

A consumer organization estimates that over a 1-year period

1919%

of cars will need to be repaired once,

66%

will need repairs twice, and

11%

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

nothing.

(Do not round.)

b) The probability that a car will require no more than one repair is

nothing.

(Do not round.)

c) The probability that a car will require some repairs is

nothing.

(Do not round.)

Ans:

Given that

P(repair once)=0.19

P(repair twice)=0.66

P(repair 3 or more)=0.11

a)P(no repair)=P(x=0)=1-P(x>=1)=1-0.19-0.66-0.11=1-0.96=0.04

b)P(x<=1)=P(x=0)+P(x=1)=0.04+0.19=0.23

c)P(some repair)=(x=1,2,3 or more)=P(x>=1)=0.19+0.66+0.11=0.96

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