Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose t

Susan Nall 2022-01-06 Answered
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?

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John Koga
Answered 2022-01-07 Author has 4059 answers

There are 8 white, 4 black and 2 orange balls. Random variable X denotes the winning amount. The values of X corresponding to otcomes of the event are: \(\displaystyle{W}{W}=-{2},{W}{O}=-{1},{O}{O}={0},{W}{B}={1},{B}{O}={2},{B}{B}={4}\)
Total number of choosing 2 balls from the urn containing 8+4+2 balls is 14C2. Number of choosing two white balls is 8C2 as they are only 8 white balls.
\(\displaystyle{P}{\left({x}=-{2}\right)}={\frac{{{n}_{{{W}{W}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{8}}}{{{2}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{28}}}{{{91}}}} \)
Number of choosing one white and one orange ball is \(\displaystyle{\left({1}{w}{h}{i}{t}{e}\right)}{\left({1}{\quad\text{or}\quad}{a}{n}ge\right)}\) as both are independent events.
\(\displaystyle{P}{\left({X}=-{1}\right)}={\frac{{{n}_{{{W}{O}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{8}}}{{{1}}}}{\frac{{{2}}}{{{1}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{16}}}{{{91}}}}\)
\(\displaystyle{P}{\left({X}={1}\right)}={\frac{{{n}_{{{W}{B}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{8}}}{{{1}}}}{\frac{{{4}}}{{{1}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{32}}}{{{91}}}}\)
\(\displaystyle{P}{\left({X}={0}\right)}={\frac{{{n}_{{{O}{O}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{2}}}{{{2}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{1}}}{{{91}}}}\)
\(\displaystyle{P}{\left({X}={2}\right)}={\frac{{{n}_{{{O}{B}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{2}}}{{{1}}}}{\frac{{{4}}}{{{1}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{8}}}{{{91}}}}\)
\(\displaystyle{P}{\left({X}={4}\right)}={\frac{{{n}_{{{B}{B}}}}}{{{n}_{{to{t}{a}{l}}}}}}={\frac{{{\frac{{{4}}}{{{2}}}}}}{{{\frac{{{14}}}{{{2}}}}}}}={\frac{{{6}}}{{{91}}}}\)

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MoxboasteBots5h
Answered 2022-01-08 Author has 3742 answers
Very accurate answer, thanks
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