One of three dice is chosen at random and is thrown three times. The only information we have is that the outcomes were 6,5,3 and that
- Dice 1 is a normal dice with giving outcome $i$ with probability $1/6$
- Dice 2 is such that it gives outcome $i$ with probability $i/21$
- Dice 3 is such that it gives outcome $i$ with probability $(2-(-1)^i)/12$
Given this information, what is the conditional probability that it was the $n:$th dice that was chosen? $n=1,2,3.$
I've solved a similar problem, but in that case they were asking about the probability that a specific card came from deck 2. There were only two decks. But here I have three dice, and I don't know how to apply Bayes' formula on this, don't I need three events?
First define some events:
$D1$: the die chosen was die $1$
$D2$: the die chosen was die $2$
$D3$: the die chosen was die $3$
Define $O$ as the event of the outcome being $6,5,3$
So, you're asked to calculate $P(D1|O)$, $P(D2|O)$, and $P(D3|O)$
And yes, now use Bayes' formula. E.g.:
$$P(D1|O) = \frac{P(O|D1) \cdot P(D1)}{P(O)}$$
where you already know that:
$$P(D1)=P(D2)=P(D3)=\frac{1}{3}$$
Also, use:
$$P(O)=P(O|D1) \cdot P(D1) + P(O|D2) \cdot P(D2) + P(O|D3) \cdot P(D3)$$
So, it's just a matter of calculating $P(O|D1)$, $P(O|D2)$, and $P(O|D3)$ and then you can plug everything in. Good luck!