Update the posterior after sampling from a mixture Suppose I put 10 red balls and 40 blue balls into a bucket. Then I put another 50 balls in that I drew at random from a set of 100 red balls and 100 blue balls.
At this point I believe the probability of drawing a red ball from the bucket is 0.35 and the probability of drawing a blue ball is 0.65. Is this correct?
Now if I actually draw a ball, and it is red, how do I calculate the probability of drawing each color on the second ball? And similarly if the first ball were blue?
This is not homework - it is an attempt to create a simple analogy from a more convoluted problem I am trying to solve.
 A: Let's call the event of drawing the first ball red $R_1$ and the event of the first ball coming from the initial fifty $F_1$. Then  
$$\Pr(R_1)=\Pr(R_1\mid F_1)\Pr(F_1)+\Pr(R_1\mid F_1^c)\Pr(F_1^c) = \frac{10}{50} \frac{50}{100} + \frac{100}{200} \frac{50}{100}  =\frac{7}{20}$$ which is $0.35$ as you suggested  
Now let's call the event of drawing the second ball red $R_2$ and the event of the second ball coming from the initial fifty $F_2$. Then 
$$\Pr(R_2 \mid R_1) = \dfrac{\Pr(R_1 \land R_2)}{\Pr(R_1)}$$ 
$$=\left(\Pr(R_1 \land R_2\mid F_1 \land F_2)\Pr(F_1 \land F_2)+\Pr(R_1 \land R_2\mid F_1 \land F_2^c)\Pr(F_1 \land F_2^c)+\Pr(R_1 \land R_2\mid F_1^c \land F_2)\Pr(F_1^c \land F_2)+\Pr(R_1 \land R_2\mid F_1^c \land F_2^c)\Pr(F_1^c \land F_2^c)\right)/\Pr(R_1)$$ 
$$=\left(\tfrac{10}{50}\tfrac{9}{49}\tfrac{50}{100}\tfrac{49}{99}+ \tfrac{10}{50}\tfrac{100}{200}\tfrac{50}{100}\tfrac{50}{99}+ \tfrac{100}{200}\tfrac{10}{50}\tfrac{50}{100}\tfrac{50}{99}+ \tfrac{100}{200}\tfrac{99}{199}\tfrac{50}{100}\tfrac{49}{99}     \right)/\tfrac{7}{20} $$
and if you do the arithmetic you will get an answer slightly less than $0.35$, which is what you might intuitively expect 
