Bayes' theorem, finding the bag the ball was chosen from, $\text{Bag}_1$ or $\text{Bag}_2$

Question

$\Pr\left(\text{Blue}|\text{Bag}_1\right) = 0.7$.

$\Pr\left(\text{Blue}|\text{Bag}_2\right) = 0.3$.

I choose balls with replacement from a bag until I have $12$ balls, $8$ of which are blue.

Therefore, the following is my working

$\Pr\left(\text{Blue}|\text{Bag}_x\right) = 0.66$.

But $\Pr\left(\text{Bag}_x|\text{Blue}\right)$ cancels out to $0.66$ using Bayes' Theorem. I'm sure that's wrong right?

Answer 1

You are using the event "Blue" in your attempt. This event is illdefined and even if you were to define it properly, it is not at all what we are interested in. Recall, the problem statement begins "given that having drawn twelve balls that eight of them are blue..." or... at least it would begin with this if you had used proper grammar when trying to copy the question.

It is possible that you just could not understand what the question was asking. Given context clues, I will assume that the correct question is the following:

We have two bags: Bag 1 and Bag 2. Bag 1 contains $7$ blue balls and $3$ red balls. Bag 2 contains $3$ blue balls and $7$ red balls. We select a bag uniformly at random and then repeatedly draw balls independently and uniformly at random with replacement from the selected bag until we have drawn $12$ in total. If given that among those twelve balls drawn there were exactly eight of them which were blue, what is the conditional probability that the selected bag happened to be the first one?

Note that what we are conditioning on is the quantity of blue balls of those twelve selected overall... not just the probability of drawing a blue if we were to draw only one ball at a time. While yes, $\Pr(\text{When drawing one ball, that one ball is blue}\mid \text{Bag}_1)=0.7$ this is very different than $\Pr(\text{When drawing twelve balls, eight of them are blue}\mid \text{Bag}_1)$.

So, let $X$ be a random variable denoting the number of the twelve balls drawn which are blue. Since the balls are drawn with replacement, we recognize that $X$ will be a binomial random variable with $n=12$ and where $p$ will vary depending on which bag we are drawing from. In the case that we drew from the first bag, $p$ will be $0.7$. In the case that we drew from the second bag, $p$ will instead be $0.3$. We have then:

$$\Pr(X=8\mid \text{Bag}_1) = \binom{12}{8}\cdot 0.7^8\cdot 0.3^4$$

$$\Pr(X=8\mid \text{Bag}_2) = \binom{12}{8}\cdot 0.3^8\cdot 0.7^4$$

$$\Pr(X=8) = \Pr(\text{Bag}_1)\cdot \Pr(X=8\mid \text{Bag}_1) + \Pr(\text{Bag}_2)\cdot \Pr(X=8\mid \text{Bag}_2)$$

The question is asking us for $\Pr(\text{Bag}_1\mid X=8)$. We approach with Bayes' Theorem and we find:

$$\Pr(\text{Bag}_1\mid X=8) = \dfrac{\Pr(X=8\mid \text{Bag}_1)Pr(\text{Bag}_1)}{Pr(X=8)}$$

Completing the necessary arithmetic finishes the problem.