i've just finished Chapter 4 of "a first course in probability 8th edition" which deals with defining Discrete Random Variables, calculating the Expectation and Variance, and gives a bit of background knowledge on common discrete random variables (the Binomial, poisson, geometric, negative binomial and hypergeometric).
my issue is the first question of the problem sheet. it goes as follows
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 \$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?
now, ive correctly determined the possible values of X and the probabilities but i feel like i was ment to utilise one of the distributions that i had just learnt about. but i couldnt figure out how as X is winnings rather than balls selected (otherwise i'd probably have used the hypergeometric distribution)
rather than my way which is as follows
we have a maximum winnings of \$4, denoted by 2 black, and a minimum of \$-2. this means that X takes values $-2,-1,0,1,2,4$ and these represent either picking 2 black balls or 1 black ball and any other ball. 2 orange balls, 2 white balls and finally 1 white ball and any other ball.
this leaves me with
$$P(X=4)=\frac{4 \cdot 3}{14 \cdot 13} = \frac{6}{91}$$ $$P(X=2)=\frac{2 \cdot 4 \cdot 2}{14 \cdot 13} = \frac{8}{91}$$ $$P(X=1)=\frac{2 \cdot 4 \cdot 8}{14 \cdot 13} = \frac{32}{91}$$ $$P(X=0)=\frac{ 2 \cdot 1}{14 \cdot 13} = \frac{1}{91}$$ $$P(X=-1)=\frac{2 \cdot 8 \cdot 2}{14 \cdot 13} = \frac{16}{91}$$ $$P(X=-2)=\frac{8 \cdot 7}{14 \cdot 13} = \frac{28}{91}$$
so my question is. was there another method to do this?
thanks in advance.
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