A country conducts a study on new cars within the first 90 days of use (Statistics)

A country conducts a study on new cars within the first 90 days of use. The cars have been
categorized according to whether the car needs a warranty–based repair (yes or no) and the car’s
origin (domestic or foreign). Based on the data collected, the probability that the new car needs
warranty repair is 0:09, the probability that the car was manufactured by a domestic company is
0:76; and the probability that the new car needs a warranty repair and was manufactured by a
domestic company is 0:009:
(a) What is the probability that a new car needs a warranty repair or was made by a domestic
company?
(b) Given that the new car was manufactured abroad, what is the probability that it does not need
a warranty repair?

Here is my answer to question a:

a) P(AorB)= P(A) + P(B) - P(A&B) = 0.09 + 0.76 - 0.009 = 0.841

b)??? P (A' and B')?????

Best answer guaranteed!

here's a simple way to understand what's going on !

take a conveniently large # of cars to avoid decimals, say 100,000
76,000 are domestic, 24,000 foreign
at Pr = 0.009, 684 domestic need repair, 75,316 don't
at Pr = 0.09, 6840 total cars need repair, so
out of 24,000 foreign, 6156 need repair, 17,844 don't

(a) (76,000+6156)/100,000 = 0.82156 <-----

(b) 17,844/24,000 = 0.7435 <------