Help with Bayes Rule and Probability

Question

I've seen many similar problems to this one, but I'm still a bit confused as to how to solve it:

Two buckets:

bucket $1$ has $3$ black balls, $7$ blue balls

bucket $2$ has $7$ black balls, $3$ blue balls

We draw randomly $12$ times from one of the buckets with replacement. This yielded $4$ black and $8$ blue.

What's the probability that the draw was made from the first bucket?

Answer 1

Hints:

Since you're just learning Bayes' Theorem, you need to break up the problem into parts. Bayes' Theorem for this problem will look like this (please verify you understand how I got this):

Let $J_i$ be the event that you are drawing from jar $i$. Let $D$ be the event that you drew 4 black and 8 blue balls.

$$P(J_i|D)=\frac{P(J_i)P(D|J_i)}{\sum_{i=1}^2 P(J_i)P(D|J_i)}$$

Hint: All you are doing is "renormalizing" your probabilities to 1, assuming you are in the part of the Venn diagram where $D$ has occurred. Draw the Venn Diagram for this problem and you'll see why.

1. What is the implicit assumption about how you pick jars? Is each jar equally likely to be chosen?
2. Can you derive the probability of getting $D$ assuming you are drawing from the first jar? Derive this for the second jar too.
3. Now, can you calculate the probability of $D$ regardless of which jar you choose (unconditional probability)?

Ok, you now have all the ingredients to solve this problem. Draw your Venn Diagram, study the above formula and Baye's Theorem and solve it. :-)

Answer 2

Let $F$ be the event of "getting$4$ black and $8$ blue",and let $E$ be the event "the selection is from bucket $1$". Thus we have:

$P\left(E\mid F\right) = \dfrac{P\left(E \cap F\right)}{P\left(F\right)}= \dfrac{P\left(E \cap F\right)}{P\left(E \cap F\right) + P\left(E^c \cap F\right)}= \dfrac{\binom{12}{4}(0.3)^4(0.7)^8}{\binom{12}{4}(0.3)^4(0.7)^8+\binom{12}{4}(0.7)^4(0.3)^8}=...$