Answer to Question #176475 in Statistics and Probability for Bikram Chaudhary

Question

If the probabilities are 0.87, 0.36, and 0.29 that, while under warranty, a new car will require repairs in the engine, drive train, or both, what is the probability that a car will require one or the other or both kinds of repairs under the warranty?

Answer 1

$P(A) = 0.87 \\ P(B) = 0.36 \\ P(A \cap B) = 0.29$

We have to find:

$P(A \cup B) = P(A) + P(B) - P(A \cup B) \\ = 0.87 + 0.36 -0.29 \\ = 0.94$