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 Dec 11, 2018
edited by shann0n  Dec 14, 2018
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\(P[6] = \dfrac{9}{10}\cdot \dfrac 1 6 + \dfrac{1}{10} \dfrac{18}{100}= \dfrac{21}{125} \)

 

\(P[\text{biased|6}] = \dfrac{P[\text{6|biased}]P[\text{biased}]}{P[6]} = \\ \large \dfrac{\frac{18}{100}\frac{1}{10}}{\frac{21}{125}} = \dfrac{3}{28}\)

 

\(P[\text{biased|!6}] = \dfrac{P[\text{!6|biased}]P[\text{biased}]}{P[\text{!6}]}=\\ \dfrac{\frac{82}{100}\frac{1}{10}}{1-\frac{21}{125}} = \dfrac{41}{416}\)

 

You can do the last one... see if you learned anything

 Dec 11, 2018

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