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 March 28th, 2018, 09:27 PM #1 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 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. Tags expectation, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jaket1 Probability and Statistics 6 November 27th, 2017 01:17 PM Jaket1 Probability and Statistics 4 October 23rd, 2017 10:59 AM eulerrules1 Advanced Statistics 6 December 9th, 2011 09:04 PM hendaz Algebra 2 May 23rd, 2010 01:57 PM hendaz Advanced Statistics 1 May 18th, 2010 01:13 PM

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