- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 4. Multiplication Rule for Independent Events

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**CFA Practice Question**

A multiple-choice test has three questions, each with four choices for the answer, of which only one of the choices is correct. What is the probability of guessing correctly on at least one of the questions?

B. 0.578

C. 0.250

A. 0.422

B. 0.578

C. 0.250

Correct Answer: B

The probability of guessing all questions wrong: 0.75 x 0.75 x 0.75 = 0.4218. The probability of guessing at least one correctly is 1 - 0.4218 = 0.5781.

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**User Contributed Comments**
10

User |
Comment |
---|---|

tenny45 |
Anyone knows how to solve it the other way around (versus subtracting from negative)? what about 0.25+(0.25*0.25)+(0.25*0.25*0.25) = 32.81% ?? This includes the probibility of guessing correctly on 1 question + on 2 questions + all 3 questions... |

akanimo |
Their solution is the neatest. To do it the other way round is really long: 1 Question correct: 1 0 0 = .25 x .75 x .75 0 1 0 = .75 x .25 x .75 0 0 1 = .75 x .75 x .25 = 3 x 0.1406 = 0.4218 2 Questions correct 1 0 1 = .25 x .75 x .25 1 1 0 = .25 x .25 x .75 0 1 1 = .75 x .25 x .25 = 3 x .0468 = 0.1404 All 3 questions correct 1 1 1 = .25 x .25 x .25 = 0.0156 Probability of at least 1 correct answer = = 0.4218 + 0.1404 + 0.0156 = 0.5778 Phew! ..... like i said other way is much faster! 100 = 110 101 111 001 011 010 |

tabulator |
Loved this question! |

SamehHassan |
Whenever you hear the "at least one" it is better to calc as: 1 -( opposite prob ) |

adenisov |
well, there is another way. Through P(1stcorrect or 2ndcorrect or 3dcorrect) P(1stcorrect or 2ndcorrect) = P(1stcorrect) + P(2ndcorrect) - P(1stcorrect)*P(2ndcorrect) etc... |

quantwannabe |
Other way is P(A or B or C) = P(A) + P(B) + P(C) - P(A)P(B) -P(B)P(C) - P(C)P(A) + P(A)P(B)P(C) = 0.25 + 0.25 + 0.25 - 0.0625 - 0.0625 - 0.0625 + 0.015625 = .5781 But I like the short cut method though. |

jaroslavb |
Intuitively, when you have 3 guestions (1,2,3), each with mutliple choice (A,B,C,D), how the probability of getting at least one answer correct (0,5781) could be more compared to situation when you have 1 questions with 2 choices, of which is one correct, i.e P=0,5 ? thanx |

dmfcrowe |
But if its at least one question right, why subtract the probabilities of when you get two right or when you get three right? Surely that is only relevant if the question is what is the probability of getting just 1 right, not AT LEAST ONE right. |

idzani |
Mad tricky. |

skyguymd |
I agree |