- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 2. Probability Function

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**CFA Practice Question**

The random variable X has the following distribution:

f(x) = [c(6 - x)]/10, x = 0,1,2,3

B. 5/9

C. 18/10

f(x) = [c(6 - x)]/10, x = 0,1,2,3

What value of c makes this a legitimate probability distribution?

A. 3/10

B. 5/9

C. 18/10

Correct Answer: B

For this to be a legitimate probability distribution, Σ p(x) = 1, so c = 5/9.

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**User Contributed Comments**
12

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

Gina |
[c(6-0)/10]+[c(6-1)/10]+[c(6-2)/10]+[c(6-3)/10]=1 18c=10 c=5/9 |

noonah |
Is there a way to do this on BAII? |

Janey |
I think you could also answer that there are 4 probablities 0,1,2,3. All probability distributions = 1, so (4+5)/9 = 1. i could be wrong tho?? |

Bibhu |
Janey u are right. Better way to answer is as provided by Gina. |

whiteknight |
so the best way to solve these kind of questions is to substitute and see ? |

steved333 |
Gina and Janey are both right. Janey's pretty good with the shortcuts, and Gina, thank you for your detailed clarification. Understanding it makes it easier to remember it... |

JanLani |
Gina's step through helped simplify and clarify it a lot. Thanks |

tschorsch |
Janey's answer does not make sense. The for probabilities are the four values of f(x) after substitution of x and c. You must solve for c. There is no short cut. |

alallstar |
yeah i don't quite get janey's solution either. |

johntan1979 |
No shortcuts, and definitely a time-waster. |

gill15 |
not a time waster and takes maybe 30 seconds.....its a discrete probability that only has values for x = 0, 1, 2 and 3. just sub in each value of x into the p(x) given...sum each one and equate to 1....just like gina did... |

minhnhut |
Disregard C and sub value in, we have (6+5+4+3)/10=9/5 The only value to make the sum 1 is 5/9 |