SOLUTION: On the campus of Central Louisiana College, a student survey found that 48%
of the students were born in Louisiana. Also, 26%
of the students were born in Louisiana and work pa
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-> SOLUTION: On the campus of Central Louisiana College, a student survey found that 48%
of the students were born in Louisiana. Also, 26%
of the students were born in Louisiana and work pa
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Question 1200288: On the campus of Central Louisiana College, a student survey found that 48%
of the students were born in Louisiana. Also, 26%
of the students were born in Louisiana and work part-time.
If a student is selected at random, find the probability that they work part-time, given that they were born in Louisiana. Found 2 solutions by math_tutor2020, ikleyn:Answer by math_tutor2020(3817) (Show Source):
Work Shown:
A = person was born in Louisiana
B = person works part time
P(A) = 0.48 is given
P(A and B) = 0.26 is also given
P(B given A) = P(A and B)/P(A)
P(B given A) = 0.26/0.48
P(B given A) = 26/48
P(B given A) = 13/24
13/24 = 0.54166666666667 approximately
Roughly 54.17% of the Louisiana-born students work part time.
They ask you about the conditional probability
P(works part-time | born in Louisiana).
By the definition, this conditional probability is the ratio
P(works part-time AND born in Louisiana)/P(born in Louisiana) = = 0.5417 (rounded). ANSWER
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