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There are$8\text{white,}4\text{black, and}2$ orange balls in the urn. In the

circumstance of pulling two balls at random. Random variable $\text{X}$represents the winning amount. The values of $\text{X corresponding}$to the results of the event are as documented.

WW,WO,OO,WB,BO,BB

The total number of paths of selecting $2$balls from the urn holding $8+4+2$balls is $14C2$. The number of paths of selecting two white balls is $8C2$as there are only $8$white balls.

$\mathrm{X}:−2,−1,0,1,2,4$

$P(X=−2)=\frac{n_{WW}}{n_{\text{total}}}=\frac{\left(\begin{array}{c}8\\ 2\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{28}{91}$

The number of routes of selecting one white and one orange ball is as both are independent occurrences

$P(X=−1)=\frac{n_{WO}}{n_{\text{total}}}=\frac{\left(\begin{array}{c}8\\ 1\end{array}\right)\left(\begin{array}{c}2\\ 1\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{16}{91}$

The number of routes of selecting both orange balls

$P(X=0)=\frac{n_{OO}}{n_{\text{total}}}=\frac{\left(\begin{array}{c}2\\ 2\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{1}{91}$

The number of routes of selecting one white and one black balls

$P(X=1)=\frac{n_{WB}}{n_{\text{total}}}=\frac{\left(\begin{array}{l}8\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{32}{91}$

The number of routes of selecting one black and one orange ball

$P(X=2)=\frac{n_{OB}}{n_{\text{total}}}=\frac{\left(\begin{array}{c}2\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{8}{91}$

The number of routes of selecting both black balls

$P(X=4)=\frac{n_{BB}}{n_{\text{total}}}=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{c}14\\ 2\end{array}\right)}=\frac{6}{91}$

Probabilities associated with succeeding values$(−2,−1,0,1,2,4)\text{are}\frac{28}{91},\frac{16}{91},\frac{1}{91},\frac{32}{91},\frac{8}{91},\frac{6}{91}$ respectively.