Distribution

Deepak Chitnis

Active Member
Subscriber
Hi David,
cant figure out the following question please help
Question: Rob conniff has encountered a difficult section on multiple choice exam. There are five question in this section and each question has three equally likely answer choices. Which of the following amount is closest to the probability that he will get three or more options correct by randomly guessing?
A.4.5%
B.16.5%
C.21%
D.79.0%
As per my knowledge in average every FRM exam there is question about binomial or poison. Please help me in these question and also tell me when we are calculating n!/(n-k)!k! which function on calculator should use nPr or nCr? and also show how if possible. Sorry for the trouble.
Thank you,
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Deepak,

This is a Binomial distribution - either he gets the correct answer or he gets the wrong answer.
Probability of getting k options correctly P(Y = k) = [n!/((n - k)!*k!] * p^k * ( 1 - p)^(n-k)
For this question k = 3, n = 5, p = 1/3 and 1 - p = 2/3
Therefore P(Y = 3) = [5!/(5! - 3!)*3!] * (1/3)^3*(2/3)^2
Using nCr we get = 10*(.03704)*(0.4444) = 16.461%
The closest answer is (B) 16.5%

Thanks!
Jayanthi
 

Deepak Chitnis

Active Member
Subscriber
Hi Jayanthi,
Thanks for the reply but the correct answer is C.21.0% that's why I got stuck, see if you can solve. Also thanks for nCr function.
Thank you,
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
I agree with @Jayanthi Sankaran but she has calculated the probability of exactly three correct answers (out of five total); i.e., P(Y=3). However, the question wants the "probability that he will get three or more options correct." So it wants P(Y=3) + P(Y=4) + P(Y=5). Using @Jayanthi Sankaran same approach, then we get P(Y=4) = 4.1% and P(Y=5) = 0.4 for a total of 21.0%. We get the same result with: Pr(X>=3) = 1 - Pr(X<3) = 1 - BINOM.DIST(number_s = 2, trials =5, probability_s = 1/3, cumulative = TRUE) = 1 - 0.790123. Thanks,
 
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