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A consumer organization estimates that over a​ 1-year period 1717​% of cars will need to be repaired​ once, 55​% will need repairs​ twice, and 44​% 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?

Respuesta :

pedrocarratore

Answer:

a) 0.74

b) 0.91

c) 0.09

Step-by-step explanation:

One repair, P(X=1) = 0.17

Two repairs, P(X=2) = 0.05

Three repairs or more, P(X≥3)=0.04

a) The probability that a car will not need repairs is 1 minus the probability of needing repairs:

[tex]P(X=0) = 1-P(X=1)-P(X=2)-P(X\geq 3)\\P(X=0) = 1-0.17-0.05-0.04\\P(X=0) = 0.74[/tex]

b) The probability that a car will need no more than one​ repair is the probability of needing zero or one repair.

[tex]P(X\leq 1)=P(X=0) +P(X=1)\\P(X\leq 1)=0.17+0.74\\P(X\leq 1)=0.91[/tex]

c) Assuming that "some repairs" mean more than one repair:

[tex]P= 1-P(X\leq 1)=1-0.91\\P=0.09[/tex]

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