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?

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:

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

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

income. 10x+1300

expenses 3x+(1800+1500)

expenses 3x+3300

to break even:

income = expense,

10x + 1300 = 3x + 3300,

solve for x.

step-by-step explanation:

mark brainliest