- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 2. Unconditional, Conditional, and Joint Probabilities

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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 will be a woman engineering major?

B. 0.0924

C. 0.6400

A. 0.0396

B. 0.0924

C. 0.6400

Correct Answer: A

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.

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**User Contributed Comments**
7

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 |

LoveIvie |
Thanks Daddykay! |

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 |