Question

An advertising agency observes that 2% of the potential buyers of a product sees its propaganda per newspaper, 20% sees that propaganda on television and 1% sees the two types of propaganda. In addition to every three who see the propaganda one buys said product and 7.9% buy and do not see the propaganda

a) What is the probability that the potential buyer bought the product if he did not see the propaganda?

b) If a potential buyer buys the product, what is the probability that he has not seen the propaganda?

Answer 1

a) P(seeing propaganda) = P(seeing propaganda on newspaper) + P(seeing propaganda on television) - P(seeing propaganda on newspaper and television)

= 0.02 + 0.20 - 0.01

= 0.21

P(not seeing propaganda) = 1 - 0.21

= 0.79

P(buy and did not see the propaganda) = P(did not see the propaganda) x P(bought the product if he did not see the propaganda)

0.079 = 0.79 x P(bought the product if he did not see the propaganda)

P(bought the product if he did not see the propaganda) = 0.079 / 0.79

= 0.1

b) Bayes' Theorem: P(A | B) = P(A & B) / P(B)

P(he has not seen the propaganda | a potential buyer buys the product) = P(buy and do not see the propaganda) / P(a potential buyer buys the product)

= 0.079 / (0.079 + 0.21x(1/3))

= 0.5302