Bayes' Theorem Question with irregular die.

Question

I'm having trouble with the following question:

Suppose we have three, six-sided dice. These are the numbers on the sides of the dice:

1. {1, 2, 3, 4, 5, 6} (regular die)

2. {2, 2, 3, 4, 5, 5} (irregular die)

3. {1, 1, 2, 3, 4, 5} (irregular die)

First, we randomly select a die. Then we roll it three times and get 2 on the first roll, 3 on the second roll and 5 on the third roll. Use Bayes’ rule to determine the chance that we were using die 2. Round your answer to four decimal places.

If we call the probability that die 2 is rolled $P(A)$, and the three rolls of {2, 3, 5} $P(B)$, then here's what I have so far:

$$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{P(B|A) \cdot (1/3)}{(2/6) \cdot (1/6) \cdot (2/6)} = 1/54$$

And then I'm stuck. Any help?

Answer 1

What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(B\mid \text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as: $$P(B)=P(B|\text{die 1})P(\text{die 1})+P(B\mid \text{die 2})P(\text{die 2})+P(B\mid \text{die 3})P(\text{die 3}).$$

Answer 2

Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=\frac26\cdot\frac16\cdot\frac26=\frac1{54}$. And by the Law of Total Probability, \begin{align*}P(B)&= P(\text{dice }1)P(B|\text{dice }1)+P(\text{dice }2)P(B|\text{dice } 2)+P(\text{dice }3)P(B|\text{dice } 3)\\ &=\frac13\cdot\frac16\cdot\frac16\cdot\frac16+\frac13\cdot\frac26\cdot\frac16\cdot\frac26+\frac13\cdot\frac16\cdot\frac16\cdot\frac16\\ &= \frac1{748}+\frac1{162}+\frac1{748} = \frac1{108} \end{align*}

So you have $$P(A|B) = \dfrac{\dfrac1{54}\cdot\dfrac13}{\dfrac1{108}}=\dfrac23\approx.6667.$$