SOLUTION: At a certain college, 55% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil enginee

Question 602109: At a certain college, 55% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil engineering.
(a) What is the probability that a randomly selected female student majors in civil engineering? Round your answer to 2 decimal places.
(b) What is the probability that a randomly selected civil engineering major is female? Round your answer to 2 decimal places

Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
55% of the students are female,

P(F) = .55

21% of the students major in civil engineering.

P(C) = .21

8% of the students both are female and major in civil engineering.

P(F&C) = .08

(a) What is the probability that a randomly selected female student majors in civil engineering?

Re-word that thinking about what's GIVEN: "What is the probability that a student majors in CE GIVEN that the student is a female".

P(C|F) = ?

Use the formula:

P(C|F) = P(C&F)/P(F) 

We have P(F) = .55 

We have P(C&F) because P(F&C) = .08 is the same thing 

P(F&C) = .08

P(C|F) = P(C&F)/P(F) = .08/.55 = .1454545455

Rounded to 2 decimal places:   .15

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(b) What is the probability that a randomly selected civil engineering major is female? Round your answer to 2 decimal places

 
Re-word that thinking about what's GIVEN: "What is the probability that a student is a female GIVEN THAT she majors in CE".

P(F|C) = ?

Use the formula swapping F and C from the previous example:

P(F|C) = P(F&C)/P(C) 

We have P(C) = .21 

We have P(F&C) = .08 

P(F|C) = P(F&C)/P(C) = .08/.21 = .380952381

Rounded to 2 decimal places:   .38

Edwin

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