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Please explain each step and write clear.

A consumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 10% will need to be repaired twice, and 5% 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?

If you own three cars, what is the probability that,

(d) neither will need repairs?

(e) all three will need to be repaired twice?

(f) at least one car will need repairs?

(g) one will need to be repaired once, one will need to be repaired twice, and one will require three or more repairs?

Respuesta :

bayosanuade

Answer: (a) 0.684

               (b) 0.884

               (c) 0.001

               (d) 0.99

Step-by-step explanation:

[tex]\\[/tex]Recall from the properties of probability

[tex]\\[/tex](i) p(x)≤1

[tex]\\[/tex](ii) p(x) ≥0

[tex]\\[/tex](iii) ∑p(x) = 1

[tex]\\[/tex]That is , if the probability of success is denoted with p and probability of failure is denoted with q , then p + q = 1

[tex]\\[/tex]Given  from the question;  

[tex]\\[/tex]In one year period

[tex]\\[/tex]Probability of a randomly chosen car to be repaired once 0.2

[tex]\\[/tex]That means the probability that it won’t be repaired once = 1 – 0.2 = 0.8

[tex]\\[/tex]Probability of a randomly chosen car to be repaired twice = 0.1

[tex]\\[/tex]That means the probability that it won’t be repaired twice = 1 – 0.1 = 0.9

[tex]\\[/tex]Probability of a randomly chosen car to be repaired three or more = 0.05

[tex]\\[/tex]That means the probability that it won’t be repaired three or more times = 1 – 0.05 = 0.95

[tex]\\[/tex](a) Probability of a randomly chosen car not to be repaired mean; it won't be repaired once, it won't be repaired twice and it won't be repaired three or more times, and you should note that "and" in probability means multiplication so,

[tex]\\[/tex]p( no repair) =0.8×0.9×0.95

[tex]\\[/tex]=0.684

[tex]\\[/tex](b) Probability of not more than one repair means the probability of no repair or the probability of one repair , which means

[tex]\\[/tex]P( no more than one repair) = p( no repair) + p ( 1 repair)

                          [tex]\\[/tex]=  0.6884 + 0.2

                  [tex]\\[/tex]= 0.884

[tex]\\[/tex](c) probability of some repair means , the probability of 1 , the probability of 2 and the probability of more than 3 repair , which implies

[tex]\\[/tex]P( some repairs ) =  p( 1 repair) x p ( 2 repairs ) x p ( s or more repair )  

 [tex]\\[/tex]= 0.2 x 0.1 x 0.05

 [tex]\\[/tex]= 0.01

[tex]\\[/tex](d) p ( neither will need repair) = 1 – p ( some repair )

                                                [tex]\\[/tex]  = 1 – 0.01

               [tex]\\[/tex]= 0.99

   

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