Popsicle ProbabilityDate: 02/27/2003 at 19:22:41 From: Ashley Subject: Probability/Combinations There are 9 popsicles: 3 orange, 3 cherry, 3 grape. There are 4 children. What is the probability that all 4 children will get the flavor of their choice? We know the answer is 26/27 but we do not understand how to get to that answer and why. We know that the first child has a 9/9 chance, the second 8/8, the third 7/7, and the forth 6/9 (2/3). We also know that there are 126 different combinations of chidren and popsicles. Date: 02/28/2003 at 14:32:59 From: Doctor Douglas Subject: Re: Probability/Combinations Hi, Ashley, Thanks for submitting your question to the Math Forum. Consider the 4 children. Each of them has a flavor preference, and so there are 3x3x3x3 = 81 possibilities for what the kids want (we assume that each kid does in fact want one of the three flavors: orange, cherry, or grape). Of these 81 possibilities, 3 of them mean that one child will be unhappy: OOOO, CCCC, and GGGG (i.e., when all four of them want the same flavor as the others). For all of the other possibilities (e.g. OOCG, OOOC, OOCC,...) there are going to be enough popsicles of the required flavors. Assuming that each of the 81 possibilities is equally likely, Pr(4 happy kids) = 1 - Pr(at least one unhappy kid) = 1 - 3/81 = 1 - 1/27 = 26/27 A supply of four of each flavor would have guaranteed four happy kids, but a supply of three of each flavor works in almost all (96%) of the cases. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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