### CFA Practice Question

There are 985 practice questions for this topic.

### 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 will be a woman engineering major?

A. 0.0396
B. 0.0924
C. 0.6400

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. An application of the general multiplication rule gives P(A and B) = P(B)P(A|B) = 0.18 x 0.22 = 0.0396.

User Comment
morek isn't this the joint multiplication rule, not the general multiplication rule?
fedha This is a difference between multiplication rule of probability (pg 325) P(AB)=P(A|B)* P(B) and Multiplication rule for Independent Events(pg 329)
P(AB)= P(A)*P(B)...
Daddykay Alternatively, assume there are 100 students total.
42% are women=42
18% engineering majors=18
of the above 18, 22% are women( 22%*18)=3.96
3.96/100=0.0396
johntan1979 P(WE)=P(W|E) x P(E)=.22 x .18 = .0396
mdejesus i used a tree diagram
the first level "branch" is .18 Eng, .82 Non Eng
the second level "branch" on "Eng" is .22 Women, .78 Men
so, following the branches "Eng" to "Women", you see .18 and .22
multiply them together (.18 x .22) to get .0396
Safiya921 P (W) = 0.42
P (E) = 0.18
P (W I E) = 0.22
P (WE) = ?

Using the formula,
P (WE) = P (W I E) P (E)
X = (0.22) (0.18)
X = 0.0396