Susan Nall

2022-01-06

Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose that we win 2 for each black ball selected and we lose 2 for each black ball selected and we lose 1 for each white ball selected. Let X denote our winnings. What are the possible values of X, and what are the probabilities associated with each value?

John Koga

Beginner2022-01-07Added 33 answers

Random variable X denotes the winning amount. The values of X corresponding to otcomes of the event are: $WW=-2,WO=-1,OO=0,WB=1,BO=2,BB=4$

Total number of choosing 2 balls from the urn 14C2. Number of choosing two white balls is 8C2 as they are only 8 white balls.

$P(x=-2)=\frac{{n}_{WW}}{{n}_{total}}=\frac{\frac{8}{2}}{\frac{14}{2}}=\frac{28}{91}$

Number of choosing one white and one orange ball is $\left(1white\right)\left(1{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}ange\right)$ as both are independent events.

$P(X=-1)=\frac{{n}_{WO}}{{n}_{total}}=\frac{\frac{8}{1}\frac{2}{1}}{\frac{14}{2}}=\frac{16}{91}$

$P(X=1)=\frac{{n}_{WB}}{{n}_{total}}=\frac{\frac{8}{1}\frac{4}{1}}{\frac{14}{2}}=\frac{32}{91}$

$P(X=0)=\frac{{n}_{OO}}{{n}_{total}}=\frac{\frac{2}{2}}{\frac{14}{2}}=\frac{1}{91}$

$P(X=2)=\frac{{n}_{OB}}{{n}_{total}}=\frac{\frac{2}{1}\frac{4}{1}}{\frac{14}{2}}=\frac{8}{91}$

$P(X=4)=\frac{{n}_{BB}}{{n}_{total}}=\frac{\frac{4}{2}}{\frac{14}{2}}=\frac{6}{91}$

MoxboasteBots5h

Beginner2022-01-08Added 35 answers

Very accurate answer, thanks

Don Sumner

Skilled2023-05-11Added 184 answers

Eliza Beth13

Skilled2023-05-11Added 130 answers

madeleinejames20

Skilled2023-05-11Added 165 answers

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