Question

A consumer organization estimates that over a 1-year period 16% of cars will need to be repaired once, 7% will need

repairs twice, and 1% will require three or more repairs. If you own two cars, what is the probability that
a) neither will need repair?
b) both will need repair?
c) at least one car will need repair?
The probability that neither will need repair is
(Do not round.)
The probability that both will need repair is
Do not round.)
The probability that at least one car will need repair is
Do not round.)

Answer 1

a) The probability that neither car will need repair is the complement of the probability that at least one car will need repair. So, P(neither car needs repair) = 1 - P(at least one car needs repair).

To find P(at least one car needs repair), we need to use the complement again. The probability that no car needs repair is:

P(no repairs) = 0.84 * 0.93 * 0.93 = 0.6961

So, P(at least one car needs repair) = 1 - P(no repairs) = 1 - 0.6961 = 0.3039

Therefore, P(neither car needs repair) = 1 - 0.3039 = 0.6961

b) The probability that both cars will need repair is:

P(both cars need repair) = P(one repair each) + P(both need two repairs) + P(one needs two, one needs three or more)

P(one repair each) = 0.16 * 0.07 = 0.0112

P(both need two repairs) = 0.07 * 0.07 = 0.0049

P(one needs two, one needs three or more) = 2 * (0.16 * 0.01) = 0.0032

Therefore, P(both cars need repair) = 0.0112 + 0.0049 + 0.0032 = 0.0193

c) The probability that at least one car will need repair is 0.3039 (calculated in part a). Therefore, the probability that both cars will need repair or at least one car will need repair is:

P(both or at least one) = P(both) + P(at least one) - P(both and at least one)

P(both and at least one) = P(both) = 0.0193 (calculated in part b)

Therefore, P(both or at least one) = 0.0193 + 0.3039 - 0.0193 = 0.2846