Normal and biased dice

Question

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?

Answer 1

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, $$\mathbf P(b\,|\,6)=\frac{0.45\cdot 0.5}{0.45\cdot 0.5+1/6\cdot 0.5}=\frac{27}{37}.$$

Answer 2

$P(6) = P(6|A)P(A)+P(6|B)P(B)+P(6|C)P(C)$

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$

Answer 3

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 $\frac{5100}{6400}= \frac{51}{64}$.