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!!
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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.
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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
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