Help with Bayes Rule and Probability 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?

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