In developing their warranty policy, an automobile company estimates that over a 1-year period 17% of their new cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you bought two cars from this company, what is the probability that neither will need repair over a 1-year period

Step-by-step explanation:

OVER A 1-YEAR PERIOD, it is estimated that

17 + 7 + 4 = 28% of new cars will need repairs (ranging from one repair to infinity number of repairs)

This means that the remaining percentage of new cars will NOT need any repairs over the 1-year period.

This percentage is 100 - 28 = 72%

The question is reframed thus: If you bought 2 cars from this company, what is the probability that none of them will need repair over the 1-year period?

What is the probability that none of the 2 cars falls within the 28% of cars which will need repair in the 1-year period?

72% = 72/100 = 0.72

this is the probability that a single car will not need repairs

the probability that 2 cars from this company will not need repair over a year is  0.72 ÷ 2 = 0.36

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

hello, there is a new name in town and that's the human spider!

answer: d.) the total number of protons and neutrons is 15

merry christmas!

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your friendly neighborhood human spider! .