 My Math Forum Two colored balls, drawing with and without replacement. Find expected number.
 User Name Remember Me? Password

 Probability and Statistics Basic Probability and Statistics Math Forum

 June 5th, 2016, 03:29 AM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 Two colored balls, drawing with and without replacement. Find expected number. Hi, I have done the following exercise but the result is different from the solution given by the book. So: The exercise: An urn containing 4 black balls and 8 white balls is used for two experiments. - In experiment 1, two balls are drawn one after another, without replacement. - In experiment 2, one ball is drawn from 12 and replaced before another ball is drawn. For each of the two experiment calculate the expected number of black balls which will be drawn. If in experiment 2, the urn contains b black balls and w white balls, where b + w = 12, calculate the expected number of black balls which will be drawn. Start: >In the experiment 1: I consider the random variable X = "number of black balls", so the r.v. can retrieve 0,1,2: - 0 black balls = P(X = 0) = P(WW) = (8/12)(7/11) = 56/132 = 14/33 - 1 black ball = P(X = 1) = P(BW)(2!) = (4/12)(8/11)(2!) = 64/132 = 16/33 - 2 black balls = P(X = 2) = P(BB) = (4/12)(3/11) = 12/132 = 3/33 here the expected number of black balls is: E(X) = (0)(14/33) + (1)(16/33) + (2)(3/33) = 16/33 + 6/33 = 22/33 >In the experiment 2: I consider the random variable Y = "number of black balls", so the r.v. can retrieve 0,1,2: - 0 black balls = P(Y = 0) = P(WW) = (8/12)(8/12) = 64/144 = 4/9 - 1 black ball = P(Y = 1) = P(BW)(2!) = (4/12)(8/12)(2!) = 64/144 = 4/9 - 2 black balls = P(Y = 2) = P(BB) = (4/12)(4/12) = 16/44 = 1/9 here the expected number of black balls is: E(Y) = (0)(4/9) + (1)(4/9) + (2)(1/9) = 4/9 + 2/9 = 6/9 = 2/3 Now, I go to calculate the probability if in the experiment 2 there are: b black balls, and w white balls. - 0 black balls = P(Y = 0) = P(WW) = (w/12)(w/12) = (w^2)/(144) - 1 black ball = P(Y = 1) = P(BW)(2!) = (b/12)(w/12)(2) = (2bw)/(144) - 2 black balls = P(Y = 2) = P(BB) = (b/12)(b/12) = (b^2)/(144) here the expected number of black balls is: E(Y) = (0)(w^2/144) + (1)((2bw)/(144)) + (2)((b^2)/(144)) = = (2bw)/(144) + (2b^2)/(144) = (2bw + 2b^2)/(144) = (2b(w+b))/144 = (b(w+b))/72 END. But my book says that the correct answers are: 2/3, 2/3, b/6 or (2b)/(b+w) Where are the errors? Are they in my book or in my way of proceeding? Please, can you help me? Many Thanks! June 5th, 2016, 08:54 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 Let's start with experiment 1. You got $\dfrac{22}{33} = \dfrac{2 * \cancel{11}}{3 * \cancel{11}} = \dfrac{2}{3}.$ Just what the book says. In experiment 2 you got (2 / 3), just what the book says. In the more general case w = number of white balls in the urn b = the number of black balls in the urn and (b + w) = 12 So everywhere you have 144 you could have $(b + w)^2$, and anywhere there is a (b + w) you could have 12. You end up with $\dfrac{2b(b + w)}{144} = \dfrac{\cancel{2}b(b + w)}{\cancel{2} * 72} = \dfrac{b(b + w)}{72} = \dfrac{b * \cancel{12}}{6 * \cancel{12}} = \dfrac{b}{6}.$ But that is the same as $\dfrac{2b(b + w)}{(b + w)^2} = \dfrac{2b * \cancel{(b + w)}}{(b + w) * \cancel{(b + w)}} = \dfrac{2b}{(b + w)} = \dfrac{2b}{12} = \dfrac{b}{6}.$ EDIT: The problem would have been better stated if it had NOT stipulated that (b + w) = 12 because THAT problem would have forced you to think about the general case, and you would not have been tempted to use 12 instead of (b + w). Thanks from beesee Last edited by JeffM1; June 5th, 2016 at 09:10 AM. Tags balls, colored, drawing, expected, find, number, replacement 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 Skyer Algebra 3 January 6th, 2014 11:26 AM simonbinxs Algebra 5 July 17th, 2012 03:12 PM butabi Advanced Statistics 1 November 27th, 2011 08:19 AM 450081592 Advanced Statistics 0 November 25th, 2011 08:16 PM frendlyfranz Algebra 8 February 19th, 2008 09:40 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      