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Phys Scientists & Engineers I

by: Quinn Larkin

Phys Scientists & Engineers I PHY 183

Marketplace > Michigan State University > Physics 2 > PHY 183 > Phys Scientists Engineers I
Quinn Larkin
GPA 3.67

Tibor Nagy

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Tibor Nagy
Class Notes
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This 1 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 183 at Michigan State University taught by Tibor Nagy in Fall. Since its upload, it has received 16 views. For similar materials see /class/207621/phy-183-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15
We are told that we have a large number of chocolates which are divided into ve avors call them a b c d and e To answer the questions we will make the following assump tions 1 There are equal numbers of each avor ie there are as many as as there are b7s etc 2 Removing 1 3 chocolates does not signi cantly affect the ratios of avors eg we pull an 7a7 on the rst sampling does not make us less likely to draw an 7a7 on the second draw For the rst question we are selecting 2 chocolates Due to the relatively small numbers involved we can write out all 25 permutations aa ab ac ad ae ba bb bc bd be ca cb cc cd ce da db dc dd de ea eb ec ed ee Since we are asked for combinations order does not matter so we cross out the duplicates above aa ab ac ad ae lea bb bc bd be ea eb cc cd ce ela elb de dd de ea eb ee ed ee and are left with just 15 combinations Alternatively we can analyze the problem this way If the rst choice is 7a7 there are 5 possibilities for the second choice a b c d e If the rst choice is 7b7 there are 4 possibilities for the second choice b c d e We eliminate 7a7 on this second round because we are looking for combinations and dont want to double count from when a was the rst choice If the rst choice is 7c7 there are 3 possibilities for the second choice c d e If the rst choice is 7d7 there are 2 possibilities for the second choice d e And if the rst choice is 7e7 there is only one possibility for the second choice e Adding the combinations in these scenarios we get 5 4 3 2 1 15


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