### CFA Practice Question

There are 985 practice questions for this topic.

### CFA Practice Question

The random variable X has the following distribution:
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

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

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