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Thread: Probability Distribution of Random Variables

  1. #1
    Mar 2010

    Probability Distribution of Random Variables

    Hi there,

    Let me begin by introducing myself. My name is Zoheb and I'm a currently studying for a BSc Computer science degree in London (my hometown). I use to love maths (back in GCSE years) but then started to see maths as a burden. Strangely, I now enjoy the mathematical side of my degree more than anything else. But enjoying it doesn't mean I'm particularly good at it. I have a lot to learn, re-teach myself and revise...

    In this process, I'm encountering many problems, some of which I solve myself and others, I don't. Often, it's because I've missed an important step or forgotten a rule, so I apologise for this question/problem as the solution may have been in front of my eyes, but even if it was, I haven't understood it, which is why I'm posting here...

    This is not a homework query or anything, it is an example I encountered during the course of study, in Schaum's Outline: Discrete Mathematics by S. Lipschutz and M. Lipson. A solution to the problem is given of course, but I don't quite understand the steps that have been taken to reach it. Before I begin, I apologise if this query has been answered many a time before. I'm not really sure what this falls under so didn't know what to search or even name this thread. Here is my problem/confusion... I have paraphrased the problem but values and procedures remain the same...

    Suppose we have a box which contains 12 items, 3 of which have defects. A sample of 3 items is selected from the box.
    The sample space $\displaystyle S$ has $\displaystyle \binom{12}{3}=220$ samples of size 3.
    Let $\displaystyle X$ denote the number of defective items selected, then $\displaystyle X$ is a random variable with range space $\displaystyle R{x}=\{0,1,2,3\}$

    So far, everything is straight forward. But my question/problem is: what steps are to be taken in order to obtain the distribution of $\displaystyle X$? A solution is given (below) as I stated, but I haven't quite worked out why it has ben done this way... Can someone please explain the methodology that is taken here?

    There are $\displaystyle \binom{9}{3}=84$ samples of size 3 with no defective items; hence $\displaystyle P(0)=\frac{84}{220}$.
    There are $\displaystyle 3\binom{9}{2}=108$ samples of size 3 containing 1 defective item; hence $\displaystyle P(1) = \frac{108}{220}$.
    There are $\displaystyle \binom{3}{2}\cdot 9=27 $samples of size 3 containing 2 defective items; hence$\displaystyle P(2)=\frac{27}{220}$.
    Lastly, there is only 1 sample of size 3 containing all 3 defective items; hence $\displaystyle P(3)=\frac{1}{220}$.

    I'm perfectly capable of calculating the probability if I'm given the binomal coefficients, but that's about it. I wouldn't know how to apply this method to a different question. Sorry, I hope my question is clear. Thank you all in advance.

    EDIT: Looking over the problem again, I believe this should've been in the Discrete Maths sub-forum, so I apologise. Also, I believe the aspect I'm having trouble with is to do with the application of combinations and permutations.
    Last edited by Khwarizmi; Mar 10th 2010 at 03:03 PM. Reason: modified latex mark-up
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