- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 1. Introduction
CFA Practice Question
You are given a discrete random variable, X, which has the following distribution: p(X=0, 1%, 2%, or 3%) = 25%, P(X > 3%) = 50%; P(X < 0) = 50%. Does this function satisfy the conditions for a probability function?
B. Yes
C. Not enough information to tell
A. No
B. Yes
C. Not enough information to tell
Correct Answer: A
The two key requirements for a probability function are that 0 <= p(x) <= 1 and that the sum of the probabilities p(x) over all possible values of X is equal to 1. This function meets the first test but fails the second test. The second test fails because the sum of the probabilities equals 4 * 0.25 + 0.5 + 0.5 = 2.0
User Contributed Comments 10
| User | Comment |
|---|---|
| guai | 0.25+0.5+0.5=1.25 > 1 |
| pranit | Even first condition is not satisfied because probability of any event cannot be negative and in the question it says P(X<0)=50%. |
| mordja | You are misreading it pranit. You can have a result that is less than zero, but you cant have a probability that is less than zero, or greater than one. What you refer to was that the result 'x' would be less than zero, not the probability. |
| thekid | DON'T Agree with "4*0.25".... Doesn't this "P(x=0,1,2 or 3)=0.25" mean that the Probability that X equals either 0,1,2 or 3 is .25 which translates to P(x=0)= .0625 and P(x=1)=.0625 and P(x=2)=.0625 and P(x=3)=.0625 {.25 dived by 4}. So, how come when they summed up the probability they got 4 TIMES .25? It should just be .25 + 0.5 + 0.5 = 1.25 OR 4*0.0625 + 0.5 + 0.5 = 1.25 PLEASE HELP!!! Thanks. |
| hardig | thekid-I think the first probablility is a 25% chance of being 0, 1, 2 or 3 = which translates into the 4*0.25 - then there is 0.5 change of being greater than 3 and a 0.5 chance of being less than 0. Therefore (0.25+0.25+0.25+0.25)+0.5+0.5=2. |
| johntan1979 | Agree with thekid. P(A or B) = P(A) + P(B) - P(AB) and in this case, P(AB) = 0, since X cannot be 1% and 2% (for example) at the same time. Therefore, P(X=0, 1%, 2%, or 3%) = 0.25, not 4x0.25 |
| dmfz | if P(X>3%) = 50% but P(X = 0,1%,2%,3%) then the obvious answer is NO, since the parameters of X are ONLY 0,1,2,3. |
| JMORALES | thekid: The question reads "p(X=0, 1%, 2%, or 3%) = 25%" The word 'OR' is the indicator that .25 belongs to each value of X, hope this helps |
| JMORALES | thekid: p(X=0, 1%, 2%, or 3%) = 25%, the word OR is the indicator that .25 belongs to each value of x |
| khalifa92 | mutually exclusive and exhaustive summed to 1 |