### CFA Practice Question

There are 410 practice questions for this study session.

### 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?

A. 0.422
B. 0.578
C. 0.250

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.

User Comment
tenny45 Anyone knows how to solve it the other way around (versus subtracting from negative)?

= 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 =
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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.