In a city, 49% of the adults are male. 24% of the adults are male and rent action-movie DVDs, and 10% of the adults are female and rent action-movie DVDs. If an adult is randomly selected for a survey, what is the probability that the adult rents action-movie DVDs, given that the adult is female?

0.20

0.37

0.43

0.59

0.15

Respuesta :

P(adult rents action-movie DVDs \ adult female) = P(adult female that rents action-movie DVDs)/P(adult female) = 0.1/0.51 = 0.20

Answer:

0.20

Step-by-step explanation:

Data:

A: the adult is a male, P(A) = 0.49

B: the adult rent action-movie DVDs

C: the adult is a female,  P(C) = 1 - P(A) = 1 - 0.49 = 0.51  

24% of the adults are male and rent action-movie DVDs, then P(A ∩ B) = 0.24

10% of the adults are female and rent action-movie DVDs, then P(C ∩ B) = 0.10

The conditional probability of B (the adult rents action-movie DVDs) given C (the adult is female) is computed as follows:

P(B|C) = P(C ∩ B)/P(C) = 0.1/0.51 = 0.2