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Mathematics 19 Online
OpenStudy (anonymous):

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? Will Fan+medal!!

OpenStudy (anonymous):

@Warriorz13

OpenStudy (warriorz13):

IDK Man very good at math @IsaiahCC100

OpenStudy (anonymous):

Thank you.

OpenStudy (warriorz13):

Np

OpenStudy (anonymous):

Well i think i might have it, @hoblos will you check my work? So 51% of the adults are female, i divided that by .10(adults who are female and rent action movies). I came up with .19607.

OpenStudy (hoblos):

suppose A: the adult rents action-movie DVDs B: the adult is female \[P(A|B) = \frac{ P(A∩B) }{ P(B) }\]

OpenStudy (hoblos):

you did 0.51/0.1 or 0.1/0.51 ?

OpenStudy (anonymous):

I think 0.1/.51

OpenStudy (hoblos):

yeah 0.1/0.51 is correct

OpenStudy (anonymous):

Thank you @hoblos

OpenStudy (hoblos):

any time :)

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