Any help please l A consuner organization estimates that ove

Any help please.

l. A consuner organization estimates that over a 1-year period. I7% of cars will need to be repaired once. T% will need repairs twice, and 4% will require three or more repairs. what is the probability that a car chosen at random will need: a. no repairs? b. at least one repair? e at most one repair? some repairs? Considering the cars that needed no repairs, as well as those needing one or more repairs, does this data constitute a valid probability distribution? c.

Solution

a)

Let P(0) be the probability of no repairs.

Thus, as probabilities should add up to 1,

P(0) + 0.17 + 0.07 + 0.04 = 1

P(0) = 0.72 [ANSWER]

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b)

P(at least one repair) = 1 - P(0)

= 1 - 0.72

= 0.28 [ANSWER]

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c)

P(at most 1 repair) = P(0) + P(1) = 0.72 + 0.17 = 0.89 [ANSWER]

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d)

If P(some repairs) = P(at least 2 repairs) [plural]

Then

P(at least 2) = P(2) + P(3 or more) = 0.07 + 0.04

= 0.11 [ANSWER]

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e)

YES, as we have set the sum of probabilities to 1 in part A. [ANSWER, YES]