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The probability distribution function of

  1. Aug 15, 2004 #1
    I've been practicing on how to get the probability distribution/density functions of certain random variables by solving some questions in my book. I cam accross this particular problem, and though, It seems easy, the answer does not comply with what I got (or simply I got the wrong answer.)

    Urn I and Urn II each has two red chips and two white chips. Two chips are drawn from each urn without replacement. Let [tex] X_1 [/tex] be the number of red chips taken from Urn I, [tex] X_2 [/tex] be the number of red chips taken from Urn II. Find the
    [tex]p_X_3(k)[/tex] where [tex]X_3 = X_1 + X_2[/tex]

    I got the answer when [tex]X_3 = 0[/tex] so thought I go with the case where [tex]X_3 = 1[/tex] and this can happen if either [tex]X_1 = 1[/tex] and [tex] X_2 = 0[/tex] or vice versa[tex]

    P(X_3 = 1) = \left ( \left (\begin{array}{cc}2 \\ 1 \end{array} \right) \left( \begin{array}{cc}2 \\ 1\end{array} \right) / \left( \begin{array}{cc}4 \\ 2\end{array} \right) \right) \right \cdot \left ( \left (\begin{array}{cc}2 \\ 0 \end{array} \right) \left( \begin{array}{cc}2 \\ 2\end{array} \right) / \left( \begin{array}{cc}4 \\ 2\end{array} \right) \right) \cdot 2[/tex]

    since you can have [tex]\left (\begin{array}{cc}2 \\ 1 \end{array} \right)[/tex] ways of getting 1 red chip and [tex]\left (\begin{array}{cc}2 \\ 1 \end{array} \right)[/tex] ways of getting the white chip out of [tex]\left (\begin{array}{cc}4 \\ 2 \end{array} \right)[/tex] ways of getting 2 chips from a set of 4 chips from Urn I and for Urn II there are 2 choose 0 ways of getting 0 red and 2 choose 2 ways of getting 2 white chips, so you multiply their probabilities, then multiply by two since the cases of Urn I and Urn II can interchange. I got a probability of 2/9, but when I referred to the answer at the appendix of the book the answer should be 2/90. Am I missing something did I misinterpret the question or is my computation wrong?
     
    Last edited: Aug 15, 2004
  2. jcsd
  3. Aug 15, 2004 #2
    There is going to be 6 ways for the first urn and 6 for the second urn, so that gives us 36 choices. Thus probability 2/90 is impossible.

    To start from the begining: If we draw a red on the first draw, the chances is 2/4, to draw a red a second time is now 1/3, thus two reds are 1/6. Similarly for two whites from the same urn. This leaves a 2/3 chance that we will draw both a red and a white from the same urn.

    To check the work we see that all cases must add to 1.

    0 Red = 1/6 from urn one, 1/6 from urn 2 = 1/36
    1 red = 2/3 from one, 1/6 from the second or visa versa: 4/18 = 2/9.
    2 Red = 2/3 X 2/3 = 4/9.
    3 red, same as 1 red = 2/9.
    4 red same as 0 red = 1/36.

    Thus checking our work we have a total of 1/36 + 2/9 +4/9 + 2/9 + 1/36 = 17/18. WHAT WENT WRONG?

    Well, there is another way you can draw two red, that is none from the first urn and two from the second, or visa versa; giving 2/36 to add to the case of 2 reds.
     
    Last edited: Aug 15, 2004
  4. Aug 15, 2004 #3
    I knew it! 2/9 was the correct answer. And those were the same answers I got from solving that problem. But for some odd reason, the appendix of the book gave these answers:

    Red = 1/6 from urn one, 1/6 from urn 2 = 1/36
    1 red = 2/3 from one, 1/6 from the second or visa versa: 4/18 = 2/90.
    2 Red = 2/3 X 2/3 = 1/20.
    3 red, same as 1 red = 2/90.
    4 red same as 0 red = 1/36.

    As you can see, the book added an extra zero to those probabilities that from 1 red to 3 red. I cant believe it. The book made a typo error ^_^;;

    thanx for the clarification on the cases btw ^_^
     
    Last edited: Aug 16, 2004
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