Problem Page At a certain college, 49% of the students are female, and 19% of the students major in civil engineering. Furthermore, 12% 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. g

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Answer with Step-by-step explanation:

Let

A=Students are female

B=Students major in civil engineering

The probability that a students are female=P(A)=[tex]\frac{49}{100}=0.49[/tex]

The probability that a student major in civil engineering=P(B)=[tex]\frac{19}{100}=0.19[/tex]

The probability that students both are female=[tex]P(A\cap B)=\frac{12}{100}=0.12[/tex]

a.We have to find the probability that a random selected female student major in civil engineering.

We have to find [tex]P(B/A)[/tex]

[tex]P(B/A)=\frac{P(A\cap B)}{P(A)}[/tex]

[tex]P(B/A)=\frac{0.12}{0.49}=0.24[/tex]

Hence, the probability that a randomly selected female student majors in civil engineering=0.24

b.We have to find the probability that a random selected civil engineering major is female.

We have to find [tex]P(A/B)[/tex]

[tex]P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.19}=0.63[/tex]

Hence, the probability that a randomly selected civil engineering major is female=0.63

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