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Homework Help
Precalculus Mathematics Homework Help
Geometric Law of Probability with Dice
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[QUOTE="AllRelative, post: 6078594, member: 540873"] [h2]Homework Statement [/h2] We have a normal 6 sided dice marked from 1 to 6. There is an equal chance to get each number at every roll. Let's put 1&2 as A type, 3&4 as B type and 5&6 as C type. We roll the dice over and over until we get a number of every type. Let X be the number of rolls. We are interested in the mean of X. [h2]Homework Equations[/h2] I see three geometric distributions in there. Let Y be a Geometric Random Variable and p the probability of success: P(Y=x) = p(1-p)^(x-1) Mean(Y) = 1/p [h2]The Attempt at a Solution[/h2] I am not sure if my reasoning is legit. I split an experiment into 3 smaller ones and added their mean. I split the problem into three steps. Let E[SUB]1[/SUB]: Number of rolls until the first type appears. Let E[SUB]2[/SUB]: Number of rolls until the second type appears. Let E[SUB]3[/SUB]: Number of rolls until the third type appears. 1) Here p = 1. The first roll to eliminates a first type. Mean(E[SUB]1[/SUB]) = 1 2) Since a first type is out, The new p for either of the other two types of appeating are 4/6= 2/3. Therefore, Mean(E[SUB]2[/SUB]) = 1 / p = 1 / (2/3) = 1.5 3) The new p is now 2/6 = 1/3. So Mean(E[SUB]3[/SUB]) = 1 / p = 1 / (1/3) = 3 We can say that the Mean(X) = 1 + 1.5 + 3 = 5.5 We wait on average 5.5 rolls before the three types apprears. [/QUOTE]
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Geometric Law of Probability with Dice
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