Bayes' Theorem Help? Problem 2.10 from Introduction to Probability by Blitzstein and Hwang asks the following:

Fred is working on a major project. In planning the project, two milestones
are set up, with dates by which they should be accomplished. This serves as a way to track Fred’s progress.


Let $A_1$ be the event that Fred completes the first milestone on time, $A_2$ be the event that he completes the second milestone on time, and $A_3$ be the
event that he completes the project on time.


Suppose that $\mathsf P(A_{j+1}\mid A_j) = 0.8$ but $\mathsf P(A_{j+1}\mid A_j^\complement) = 0.3$ for $j \in\{ 1, 2\}$, since if Fred falls behind on his schedule it will be hard for him to get caught up.


Also, assume that the second milestone supersedes the first, in the sense that once we know whether he is on time in completing the second milestone, it no longer matters what happened with the first milestone. We can express this by saying that $A_1$ and $A_3$ are conditionally independent given $A_2$ and they’re also conditionally independent given $A_2^\complement$.


(a) Find the probability that Fred will finish the project on time, given that he completes the first milestone on time. Also find the probability that Fred will finish the project on time, given that he is late for the first milestone.


(b) Suppose that $\mathsf P(A_1) = 0.75$. Find the probability that Fred will finish the project on time.

My thinking is that you can obviously start with $\mathsf P(A_3 \mid A_1)$, but then you can try conditioning on $A_2$ as well. This leads me to a roadblock. Thoughts?
Yes, you do seek $\mathsf P(A_3\mid A_1)$, as well as $\mathsf P(A_3\mid A_1^\complement)$ and the next step is to use partitioning on $A_2$, with the Law of Total Probability.

$$\mathsf P(A_3\mid A_1) ~=~ \mathsf P(A_3\mid A_1,A_2)\;\mathsf P(A_2\mid A_1)+\mathsf P(A_3\mid A_1,A_2^\complement)\;\mathsf P(A_2^\complement\mid A_1)$$

$$\mathsf P(A_3\mid A_1^\complement) ~=~ \mathsf P(A_3\mid A_1^\complement,A_2)\;\mathsf P(A_2\mid A_1^\complement)+\mathsf P(A_3\mid A_1^\complement,A_2^\complement)\;\mathsf P(A_2^\complement\mid A_1^\complement)$$

So, moving onwards. &nbsp; The assumption of conditional independence, between $A_1,A_3$ when given $A_2$, would mean that: 

$$\mathsf P(A_3\mid A_1,A_2)=\mathsf P(A_3\mid A_2)\\\mathsf P(A_3\mid A_1,A_2^\complement)=\mathsf P(A_3\mid A_2^\complement)\\\mathsf P(A_3\mid A_1^\complement,A_2)=\mathsf P(A_3\mid A_2)\\\mathsf P(A_3\mid A_1^\complement,A_2^\complement)=\mathsf P(A_3\mid A_2^\complement)$$

Put it together.



Of course, for the second part we likewise use that: $$\mathsf P(A_3)~=~\mathsf P(A_3\mid A_1)\;\mathsf P(A_1)+\mathsf P(A_3\mid A_1^\complement)\;\mathsf P(A_1^\complement)$$
Solution of (a)



Solution of (b)