Bayesian Conditional Probability Problem

Question

I have the following problem: An email is sent from computer $A$ to computer $B$. The connection between the two computers is of this type, each message from $A$ is sent to $C$ and $D$, which send the message to $B$. At each step the message can arrive intact or not intact (the check is done through a code), in the case in which the message is intact is sent back according to the scheme described above, otherwise the message is deleted and lost. The probability that from one passage to another the message arrives intact is described as follows ($\mathrm{P}(A \rightarrow C)$ indicates the probability that the message sent by $A$ will reach intact to $C$ and so on): $$\mathrm{P}(A \rightarrow C) = 0.78,\mathrm{P}(A \rightarrow D) = 0.9, \mathrm{P}(C \rightarrow B | A \rightarrow C) = 0.79, \mathrm{P}(D \rightarrow B | A \rightarrow D) = 0.79, \mathrm{P}(C \rightarrow B | (A \rightarrow C)^c) = 0, \mathrm{P}(D \rightarrow B | (A \rightarrow D)^c) = 0 $$ At last, all events concerning a transmission chain ($A \rightarrow C \rightarrow B$) are stochastically independent from those of the other chain ($A \rightarrow D \rightarrow B$). The questions are:

• What's the probability that the message will be delivered ONLY through the channel $A \rightarrow C \rightarrow B$?
  
  Not sure if it's $\mathrm{P}((A \rightarrow C) \cap (C \rightarrow B)) \cup ((A \rightarrow D)^c)\cap(D \rightarrow B)^c)) = \mathrm{P}(A \rightarrow C)\cdot \mathrm{P}( C \rightarrow B | A \rightarrow C) + \mathrm{P}(A \rightarrow D)^c\cdot \mathrm{P}( (D \rightarrow B)^c | (A \rightarrow D)^c) = 0.78\cdot0.79+0.1\cdot1 = 0.7162$ (here I used the Bayes Theorem and found the $\mathrm{P}( (D \rightarrow B)^c | (A \rightarrow D)^c) = 1 - \mathrm{P}( (D \rightarrow B) | (A \rightarrow D)^c)$)
• What is the probability that the message will be delivered through the channel $A \rightarrow C \rightarrow B$?
  
  I think this is simply $\mathrm{P}((A \rightarrow C) \cap (C \rightarrow B)) = \mathrm{P}(A \rightarrow C)\cdot \mathrm{P}( C \rightarrow B | A \rightarrow C) = 0.78 \cdot 0.79 = 0.6162$ (I used Bayes again)
• What is the probability that the message will be delivered?
  
  I think this is the $\mathrm{P}((C \rightarrow B | A \rightarrow C) \cap \mathrm{P}(D \rightarrow B | A \rightarrow D)) = 0.79\cdot0.79 = 0.6241$
• Knowing that the message was received by B, what is the probability of the event $A \rightarrow D$?
  
  This is where I have a lot of doubts, because I can write this as the $\mathrm{P}(A \rightarrow D | ((C \rightarrow B) \cap (D \rightarrow B)) = $ (not sure of the following passage) = $\frac{\mathrm{P}((A \rightarrow D) \cap (C \rightarrow B) \cap (D \rightarrow B))}{\mathrm{P}((C \rightarrow B) \cap (D \rightarrow B)))} = \frac{\mathrm{P}(C \rightarrow B) \cdot \mathrm{P}(A \rightarrow D) \cdot \mathrm{P}(D \rightarrow B | A \rightarrow D))}{\mathrm{P}(C \rightarrow B) \cdot \mathrm{P}(D \rightarrow B)} = \frac{0.9 \cdot 0.79}{\mathrm{P}(D \rightarrow B)} $
  
  where the first term of both numerator and denominator are semplified but I don't know how to find the remaining denominator. Actually I'm not ever sure what I did so far is right or not. Any hints fot this problem?

Answer 1

For compactness, we'll omit the arrows with the understanding that $ACB$ is the event of the message delivered through route $A$ to $C$ to $B$, and such.

Then (1) should use: $$\mathsf P({ACB}\cap{ADB}^\complement)~{=\mathsf P(AC)\mathsf P(CB\mid AC)\;(\mathsf P(AD^\complement)+\mathsf P(AD)\mathsf P(DB^\complement\mid AD))\\=\mathsf P(AC)\mathsf P(CB\mid AC)\;(1-\mathsf P(AD)\mathsf P(DB\mid AD))}$$

Reason 1: You seek the intersection of events for passing through the one route and not the other. The two routes are independent.

Reason 2: The message will not pass through route $ADB$ if it fails at $AD$ or passes $AD$ but fails at $DB$.   Alternately, just not pass through the route.

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(2) is okay.

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(3) The message passes through either route if it passes through one route, or doesn't but passes through the other. These are disjoint options. So $ACB\cup ADB = ACB\cup (ADB\cap ACB^\complement)$ $$\mathsf P(ACB\cup ADB)= \mathsf P(AC)\mathsf P(CB\mid AC)+\mathsf P(AD)\mathsf P(DB\mid AD)(1-\mathsf P(AC)\mathsf P(CB\mid AC))$$

Reason: You seek the union of events for passing through either route. The routes are not mutually exclusive events (not disjoint).

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For (4) $AD\cap(ACB\cup ADB)= AD\cap ((DB^\complement\cap ACB)\cup DB)$

$$\mathsf P(AD\mid ACB\cup ADB) ~{=\dfrac{\mathsf P(AD)\mathsf P(ACB\cup ADB\mid AD)}{\mathsf P(ACB\cup ADB)} \\= \dfrac{\mathsf P(AD)\big(\mathsf P(DB^\complement\mid AD)\mathsf P(AC)\mathsf P(CB\mid AC)+\mathsf P(DB\mid AD)\big)}{\mathsf P(ACB\cup ADB)}}$$

Answer 2

I'm only going to address the first part, which you did incorrectly. After you fix that, you should review the rest of your work.

First the event that only the channel $A\to C\to B$ succeeds is the event that this channel succeeds, AND that the channel $A\to D\to B$ fails. So, the first $\cup$ in your formula ought to be a $\cap.$ Also, the channel $A\to D\to B$ fails if $A\to D$ fails OR if $A\to D$ succeeds and $D \to B$ fails. Therefore, the expression $$(A \rightarrow D)^c\cap(D \rightarrow B)^c$$ in your formula ought to be $$(A \rightarrow D)^c\cup((A\to D) \cap (D\rightarrow B)^c)$$

It doesn't make sense to talk about $D\ B$ in isolation, because this transmission isn't event attempted if the transmission $A \to D$ succeeds.

I think it's a lot easier to do this is you write$$ P(A\to C\to B \wedge \neg(A\to D \to B)=\\ P(A\to C)P(C\to B |A\to C)(1-P(A\to D)P(D\to B|A\to D))$$

Just the probability that the first channel succeeds times the probability that the other fails.