Question involving Bayes' Rule and the Law of Total Probability

Question

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?

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:

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

Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 22, p 87. Then the author uses Bayes' Rule to find $P(A|B)=\frac{2}{3}$.

The main question I have is where $P(C|B,A)P(A|B)+P(C|B,A^c)P(A^c|B)$ comes from. It looks like the law of total probability but I'm not quite sure how it's derived. Could anyone point me in the right direction or link me to a derivation?
Furthermore, why do we condition on $B$ here: $P(C|B,A)P(A|B)+P(C|B,A^c)P(A^c|B)$ ?

Instead of just $P(C|B,A)P(A)+P(C|B,A^c)P(A^c)$ ?

See https://math.stackexchange.com/a/3433607. – Dec 23 '21 at 07:21 — , Dec 23 '21 at 07:21

score 2 · Accepted Answer · answered May 22 '16 at 09:46

The reason why the total probability is conditioned on $A \mid B$ rather than $A$ is because in the description of the problem, event $B$ was observed with certainty: therefore, all other events, even the event that the initial marble is green, should be conditioned on $B$.

Another way to reason about this problem is to define a new set of events: $$A' = A \mid B, \quad C' = C \mid B,$$ which respectively describe the event that the initial marble is green given that a green marble was drawn; and the event that the remaining marble is green given that a green marble was drawn. Then the law of total probability gives $$\Pr[C'] = \Pr[C' \mid A'] \Pr[A'] + \Pr[C' \mid \bar A'] \Pr[\bar A'].$$ We can't condition on $A$ itself because once we observed the event $B$, we have in some sense additional information about the initial marble's color; in turn, this must be reflected in the probability of the remaining marble being green. For it would be obvious that had the marble drawn not been green, the remaining marble would be green with certainty and the initial marble was just as certainly green, rather than green or blue with an equal chance. The calculation of $\Pr[A \mid B]$ via Bayes' rule formally shows this.

How is $P(C|B,A)$ equivalent to $P(C'|A')$? Isn't $P(C'|A')=P((C|B)|(A|B))$. — jlcv, May 22 '16 at 19:38
@ZoomBee Try drawing a Venn diagram with events $A$, $B$, and $C$, and then figure out what $C \mid (A \cap B)$ means, and what $(C \mid B) \mid (A \mid B)$ means. — heropup, May 22 '16 at 20:31
@heropup You are wrong to write $C \mid (A \cap B), (C \mid B) \mid (A \mid B)$? 1. Michael Hardy wrote "There's no such thing as $A∣B$. When one writes $Pr(A∣B)$, one is NOT writing about the probability of something that's called $A∣B$." 2. What do you mean by $(C \mid B) \mid (A \mid B)$? — , Dec 23 '21 at 07:24

Question involving Bayes' Rule and the Law of Total Probability

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