Question

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?

Answer 1

Let \(M \)= male,
\(A\) = rent action movies
\(\overline{M}\) = female

You are given:
\(P(M) = 0.49 \)
\(P(M \cap A )=0.24\)
\(P(\overline{M} \cap A )= 0.1\)
And the question asks for:
\(P(A | \overline{M})\)

Use the law of conditional probability:
\[P(A|\overline{M})=\frac{P(A \cap \overline{M})}{P(\overline{M})} \]

Answer 2

And notice that \(P(\overline{M})=1-P(M)\) (the complement)
and clearly \(P(A \cap \overline{M})=P(\overline{M} \cap A\))