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March 28th, 2018, 09:27 PM   #1
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Expectation question

I have this question with my work. Could anyone have a look, thank you.
Q1)
A bucket contains 4 red balls, 3 greens balls, and 5 yellow balls. Sally will select 3 balls without replacement. Let X be the number of red balls and Y be the number of yellow balls. Find the expected value of Z=X+Y using the Finite Expectation Formula.
$$Z \sim \text{Hypergeometric}(N=12, n=3, p_Z=\frac{9}{12}).$$ $$E(Z)=\sum\limits_{Z\in R_Z}^{}zP_Z(z)=\sum\limits_{z=0}^{3}\dfrac{\dbinom {9}{z}\dbinom{3}{3-z}}{\dbinom{12}{3}}$$
$$=0\cdot \dfrac{\dbinom{9}{0}\dbinom{3}{3}}{\dbinom{12}{3}} + 1\cdot\dfrac{\dbinom{9}{1}\dbinom{3}{2}}{\dbinom{1 2}{3}} + 2\cdot\dfrac{\dbinom{9}{2}\dbinom{3}{1}}{\dbinom{1 2}{3}} + 3\cdot\dfrac{\dbinom{9}{3}\dbinom{3}{0}}{\dbinom{1 2}{3}} = 2.25$$

Q2)
A bucket contains 4 red balls, 3 greens balls, and 5 yellow balls. Sally will select 2 balls without replacement. Let X be the number of red balls and Y be the number of yellow balls. Find the expected value of Z=X+Y using the linearity property of expectations.
$$X \sim \text{Hypergeometric}(N=12, n=2, p_X=\frac{4}{12}).$$
$$Y \sim \text{Hypergeometric}(N=12, n=2, p_Y=\frac{5}{12}).$$
$$E(Z)=E(X+Y)=E(X)+E(Y)=n\times p_X+n\times p_Y = 2\cdot\frac{4}{12} + 2\cdot\frac{5}{12}=1.5$$

Last edited by Shanonhaliwell; March 28th, 2018 at 09:30 PM.
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