Why does conditioning over a conditional probability give this equation (from Harvard's STAT 110 problem set)?

Question

I have two questions from Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 22, p 87.

22. A bag contains one marble which is either green or blue, with equal probabilities. A green marble is put in the bag (so there are 2 marbles now), and then a random marble is taken out. The marble taken out is green. What is the probability that the remaining marble is also green?

The author's solution:

Let A be the event that the initial marble is green, B be the event that the removed marble is green, and C be the event that the remaining marble is green. We need to find P(C|B). There are several ways to find this; one natural way is to condition on whether the initial marble is green:

$$P(C|B) = P(C|B,A)P(A|B) + P(C|B,A^c)P(A^c|B)=1P(A|B)+0P(A^c|B)$$

We know that:

$$P((C|B)|A) = P(C|B,A)$$

and through law of total probability:

$$P(X) = \sum_i^n P(X|Y_i)P(Y_i)$$

1. But I'm struggling to see how he derived the above equation. Specifically the $P(A|B)$ in $P(C|B,A)P(A|B)$. Shouldn't it be $P(C|B,A)P(A,B)$?

2. Then the author derives $P(A|B)$ via Bayes' rule:

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{\frac{1}{2}}{P(B|A)P(A) + P(B|A^c)P(A^c)} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{4}} = \frac{2}{3}$$

The part I can't grasp is why $P(B|A^c)P(A^c) = \frac{1}{4}$. Shouldn't it be $\frac{1}{2}$ because probability of removed marble is green, given initial is blue, is 1 (since we can't have two blues), so we're just left with $P(A^c)$ which is $\frac{1}{2}$?

Answer 1

Regarding $P(C \cap A \mid B) = P(C \mid B, A) P(A \mid B)$.

• One way to see this is to just note that it is a generalization of $P(C \cap A) = P(C \mid A) P(A)$ where the probability here is not the unconditional probability, but the conditional probability given $B$. Note that by adding "conditioned on $B$" to all three terms gives you $P(C \cap A \mid B) = P(C \mid A, B) P(A \mid B)$.
• If that is not convincing enough, you can always fall back on the definitions. Just subsititute $P(C\cap A \mid B) = \frac{P(A \cap B \cap C)}{P(B)}$ and $P(C \mid B, A) = \frac{P(A \cap B \cap C)}{P(A \cap B)}$ and $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ to immediately see why the equation holds.

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$P(B \mid A^c) = \frac{1}{2}$ because if $A^c$ occurs, then the bag contains one green and one blue marble, so the [conditional] probability of taking out a green marble is $1/2$.