CFA Practice Question

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CFA Practice Question

At a certain university, 42% of the students are women and 18% are engineering majors. Of the engineers, 22% are women. If a student at this university is selected at random, what is the probability that the selected person is a woman or an engineering major?

A. 0.0396
B. 0.5604
C. 0.6000

Let A be the event that a woman is selected and B be the event that an engineering major is selected. We are given the information that P(A) = 0.42, P(B) = 0.18, and P(A|B) = 0.22. According to the general addition rule, P(A or B) = P(A) + P(B) - P(A and B).
Since P(A and B) = P(B) x P(A|B) = 0.18 x 0.22 = 0.0396
P(A or B) = 0.42 + 0.18 - 0.0396 = 0.5604.

User Comment
Gina why not simply: P(AorB)= 0.42 + 0.18 - 0.22=0.38 p(women)+p(engineers)-p(women engineers) since women engineers is the overlapping category we have counted twice. ?
Gina ignore the above. obviously it cannot be less than 42%, in the first place. 22% of 18% = 3.96% = p(women engineers) that have been counted twice.
wundac thsnkd gina. I had the same thought
fedha P(W)=0.42 P(E)=0.18 P(W|E)=0.22 P(WorE)=?

We know that P(WE)= P(W|E)* P(E)
Therefore (0.22*0.18) = 0.0396

We also know P(WorE) = P(W)+P(E)?P(WE)
Therefore 0.42 + 0.18 - 0.0396 = 0.5604
morek Thank you fedha. That is the best way to calculate it.
jgraham6 I was an engineering major in college. Trust me, there were hardly any women in my classes. I should've been a child psychology major.
2014 In that u wud have married womens in ur clinic

But withs their kids, and probability of their husbands too.

Keyword in this question is women or engineering major in earlier question they refered only to eng women.

So clue to remember this question: u never get same women in ur life again. so why to double count them