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Oscar has lost his dog in either forest A (with a priori probability 0.4) or in forest B (with a priori probability 0.6).

If the dog is alive and not found by the $N^{th}$ day of the search, it will die that evening with probability $N/(N + 2)$.

If the dog is in A (either dead or alive) and Oscar spends a day searching for it in A, the conditional probability that he will find the dog that day is 0.25. Similarly, if the dog is in B and Oscar spends a day looking for it there, he will find the dog that day with probability 0.15.

The dog cannot go from one forest to the other. Oscar can search only in the daytime, and he can travel from one forest to the other only at night.

Let $A$ be the event that the dog was lost in forest A and $A^c$ be the event that the dog was lost in forest B.
Let $D_n$ be the event that the dog dies on the $n^{th}$ day.
Let $F_n$ be the event that the dog is found on the $n^{th}$ day.
Let $S_n$ be the event that Oscar searches forest A on $n^{th}$ day and $S_n^c$ be the event that he searches forest B on day $n$.

Given that Oscar looked in A on the first day but didn’t find his dog, what is the probability that the dog is in A?

I found an answer: $Pr(A|S_1,F_1^c)=\frac{Pr(A)Pr(F_1^c|A,S_1)}{Pr(A)Pr(F_1^c|A,S_1)+Pr(A^c)Pr(F_1^c|A^c,S_1)} = \frac{0.3}{0.3+0.6}$

They indicate $Pr(F_1^c|A^c,S_1) = 1$

Consider the tree (I'm not able to upload images, yet):

|        |-0.25-F
|-0.4-A  |
|        |-0.75-F^c
|
|        |-0.15-F
|-0.6-A^c|
|        |-0.85-F^c

$Pr(A)Pr(F_1^c|A,S_1) = 0.4*0.75 = 0.3$
why is it not the case that:
$Pr(A^c)Pr(F_1^c|A^c,S_1) = 0.6*0.85 = .51$

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(apologies, I can't comment under your post) Am I correct in understanding your confusion is solely around $$Pr(F_1^c|A^c,S_1)$$? They have 1 and you have .85. I can clear that up quickly. If the dog is in B and Oscar searches in A, he definitely will not find the dog because he's looking in the wrong forest. $$Pr(F_1^c|A^c,S_1)$$ is the chance that he doesn't find the dog given that the dog is in B and he searches in A.

On to the denominator (your question in the comments). Let's review Bayes' Theorem. $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$. So in our example we want to to capture the probability the dog was not found on day one (i.e. $P(F_i^c)$). We do this by defining a set of events that fully the cover the universe of possibilities and condition this value on them (Basically utilizing $P(X) = P(X|Y)P(Y) + P(X|Y^c)P(Y^c)$). We utilize the fact that we are told $S_i$ happened. So the chance that we didn't find the dog can be broken into the chance the dog was in A and we didn't find it in A, the chance the dog was in B and we didn't find it in A (the two terms where we looked in B are skipped because $P(S_i^c)=0$ so $P(x|S_i^c) = 0$ for any event x).

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  • $\begingroup$ OK, thank you. Can you explain why the second term in the denominator is $Pr(F_1^c|A^c,S_1)$ rather than $Pr(F_1^c|A^c,S_1^c)$? $\endgroup$ – Albert Han Feb 10 '17 at 22:23
  • $\begingroup$ Edited to include this explanation $\endgroup$ – David Winton Feb 10 '17 at 22:44

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