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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?

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?

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