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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