# There is a box with 12 dice which all look the same. However there are actually three types of dice:

There is a box with $12$ dice which all look the same. However there are actually three types of dice:
$6$ normal dice. The probability to get a $6$ is $1/6$ for each dice.
$3$ biased dice. The probability to get a $6$ is $0.85$.$3$ biased dice. The probability to get a $6$ is $0.05$.You take a die from the box at random and roll it.What is the conditional probability that it is of type $b$, given that it gives a $6$?
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$P\left(6\right)=P\left(6|A\right)P\left(A\right)+P\left(6|B\right)P\left(B\right)+P\left(6|C\right)P\left(C\right)$
That is, the chance of rolling a six is the chance of rolling a $6$ on dice $A$ times the chance of choosing dice $A$. And then same for $B$, and for $C$
$P\left(6\right)=\frac{1}{6}\frac{1}{2}+\frac{17}{20}\frac{1}{4}+\frac{1}{20}\frac{1}{4}=\frac{37}{120}$
What is the chance that you rolled dice $B$ given that you rolled a $6$?
$P\left(B|6\right)=$ $\frac{P\left(6|B\right)P\left(B\right)}{P\left(6\right)}\phantom{\rule{0ex}{0ex}}\frac{\frac{17}{20}\frac{1}{4}}{\frac{37}{120}}=\frac{51}{74}\approx 0.70$