Question

2.111 (4 points) An art dealer receives a shipment of five old paintings from abroad, and, on the basis of past experience, she feels that the probabilities are, respectively, 0.76, 0.09, 0.02, 0.01, 0.02, and 0.10 that 0, 1, 2, 3, 4, or all 5 of them are forgeries. Since the cost of authentication is fairly high, she decides to select one of the five paintings – 2– MATH-4751/6751, Fall-2020 Homework #3 at random and send it away for authentication. If it turns out that this painting is a forgery, what probability should she now assign to the possibility that all the other paintings are also forgeries?

Answer 1

Answer:

0.675675675 = 67.57%

Step-by-step explanation:

P(P₀) = 0.76

P(P₁) = 0.09

P(P₂) = 0.02

P(P₃) = 0.01

P(P₄) = 0.02

P(P₅) = 0.1

P(F) is the probability that the first painting is a forgery

P(F | P₀) = 0

P(F | P₁) = 0.2

P(F | P₂) = 0.4

P(F | P₃) = 0.6

P(F | P₄) = 0.8

P(F | P₅) = 1

P(P₅ | F) = [P(F | P₅) x P(P₅)] / {[P(F | P₁) x P(P₁)] + [P(F | P₂) x P(P₂)] + [P(F | P₃) x P(P₃)] + [P(F | P₄) x P(P₄)] + [P(F | P₅) x P(P₅)]}

P(P₅ | F) = (1 x 0.1) / [(0.2 x 0.09) + (0.4 x 0.02) + (0.6 x 0.01) + (0.8 x 0.02) + (1 x 0.1)] = 0.1 / (0.018 + 0.008 + 0.006 + 0.016 + 0.1) = 0.1 / 0.148 = 0.675675675 = 67.57%