Question

Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events A1, A2, and A3 by A1 = likes vehicle #1 A2 = likes vehicle #2 A3 = likes vehicle #3. Suppose that P(A1) = 0.55, P(A2) = 0.65, P(A3) = 0.70, P(A1 ∪ A2) = 0.80, P(A2 ∩ A3) = 0.50, and P(A1 ∪ A2 ∪ A3) = 0.84. (a) What is the probability that the individual likes both vehicle #1 and vehicle #2? (b) Determine P(A2 | A3). (Round your answer to four decimal places.) P(A2 | A3) =

Answer 1

a. Use the inclusion/exclusion principle.

$P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=0.55+0.65-0.80=0.40$

b. By definition of conditional probability,

$P(A_2\mid A_3)=\dfrac{P(A_2\cap A_3)}{P(A_3)}=\dfrac{0.50}{0.70}\approx0.7143$