# Probability Distribution of random variable X.

• Nov 29th 2009, 01:07 PM
diddledabble
Probability Distribution of random variable X.
I need help. I am trying to find the Probability Distribution of random variable X. X is the score of an test. In a multiple choice test of 10 questions with k possible answers (only one is correct). Each correct answers is worth five points. There are no penalties for wrong answers. I know it needs to be the Binomial distribution but what is throwing me off is that the random variable X is the score and not the number of correct answers, even though the two are related. Any idea on how to set this up? I tried $P(X=x)= (n choose x) (1/k)^x(1-(1/k)^{(n-x)}$ but I don't think that accounts for everything.
• Nov 30th 2009, 01:38 AM
Moo
Hello,

What differs from the score and the number of right answers is that each right answer gives 5 points.
So the only possible values for x are 0,5,10,15,...,5n
Note that n=10 here.

That's all folks !
• Nov 30th 2009, 03:41 AM
diddledabble
So I need to evaluate http://www.mathhelpforum.com/math-he...80a03652-1.gif for X=0,5,10,15,...,5n but I don't know anything about k except it is greater than or equal to 2. So how do I take that into account? Should it just be a generalized formula for the probability distribution of random variable X?
• Nov 30th 2009, 03:46 AM
Moo
Yes, I think it's a general formula.
There's no way to know what k is.
• Dec 2nd 2009, 08:44 AM
diddledabble
Moo,
I still don't think that accounts for everything. You see if x takes on the values 0,5,10...5n then in the formula n choose x you would end up with a (20 choose 50) Having the X be the score seems to make it messier. Our formula if you let x take on the values 0-10 for the number of correct answers would work. But the score messes that up. I was thinking could you do something where P(X=x) is the sum over all possible y's of P(X=x & Y=y)? Where P(Y=y) is our formula with y=0,1,...10.