Normal and biased 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?

 A: $P(6) = P(6|A)P(A)+P(6|B)P(B)+P(6|C)P(C)$
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
$P(6) = \frac 16 \frac 12 + \frac {17}{20} \frac 14 + \frac {1}{20} \frac 14 = \frac {37}{120}$
What is the chance that you rolled dice B given that you rolled a 6?
$P(B|6) = $$\dfrac {P(6|B)P(B)}{P(6)}\\
\dfrac {\frac {17}{20} \frac 14}{\frac {37}{120}}= \frac {51}{74}\approx 0.70$
A: Imagine rolling each of those dice 6000 times.  Type a will come up "6" 1000 times.  Type b will come up "6" 5100 times.  Type c will come up "6" 300 times.  
So we have a total of 1000+ 5100+ 300= 6400 times when "6" came up. 5100 of those times, we were using "type b" dice.  The probability we rolled type b dice, given that we got "6" is $\frac{5100}{6400}= \frac{51}{64}$.
A: Accordingly to Bayes rule, 
$$\mathbf P(b\,|\,6)=\frac{\mathbf P(6\,|\,b)\cdot \mathbf P(b)}{\mathbf P(6\,|\,b)\cdot \mathbf P(b)+\mathbf P(6\,|\,n)\cdot \mathbf P(n)},$$
where "$n$" denotes the normal type. Clearly, $\mathbf P(6\,|\,n)=1/6$, $\mathbf P(n)=1/2$, and 
$$\mathbf P(6\,|\,b)=0.5\cdot 0.85+0.5\cdot 0.05=0.45.$$
Therefore,
$$\mathbf P(b\,|\,6)=\frac{0.45\cdot 0.5}{0.45\cdot 0.5+1/6\cdot 0.5}=\frac{27}{37}.$$
