A consumer organization estimates that over a 1-year period 18% of cars will need to be repaired once, 7% will need repairs twice, and 3% 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?

(a) 0.5184

(b) 0.0784

(c) 0.4816

Step-by-step explanation:

According to the given estimation over a 1-year period, 18% of cars will need to be repaired once, 7% will need repairs twice, and 3% will require three or more repairs.

So, the probability that a car need repair once = 18/100=0.18 ...(i)

The probability that a car needs repair twice = 7/100=0.07 ...(ii)

The probability that a car needs repair thrice or more = 3/100=0.03 ...(iii)

From equations (i), (ii), and (iii), the probability that a car needs repair = (0.18+0.07+0.03) = 0.28 ...(iv)

The probability that a car doesn't needs repair = 1-(0.18+0.07+0.03)= 1-0.28=0.72 (v)

Given that I own two cars, denoting the cars by A and B.

(a) The probability that neither will need repair=(Probability that car A doesn't need repair) and (Probability that car B doesn't need repair)

=0.72 x 0.72 [ from equation (v)]

=0.5184

(b) The probability that both will need repair = (Probability that car A needs repair) and (Probability that car B needs repair)

=0.28 x 0.28 [ from equation (iv)]

=0.0784

(c) The probability that at least one car will need repair= 1 - (The probability that neither will need repair)

=1-0.5184 [ by using part (a)]

=0.4816

the answer is c

step-by-step explanation: