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I am trying out the first probability problem at this link. I found the notations given too complicated, so I follow my own.

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To simplify the problem's wording, the probability of the dog being alive on day n given it is alive on day n - 1 is:

P(alive, day n) = P(alive, evening n - 1)
= 1 - P(dead, evening n - 1)
= 1 - (n - 1) / (n - 1 + 2)
= [(n + 1) - (n - 1)] / (n + 1)
= 2 / (n + 1)

For question f), this is my reasoning:

P(find dog, in B, day 4, alive) = P(find dog, in B) * P(day 4, alive)
= P(find dog, in B) * P(day 4, alive)
= P(find dog | in B) * P(in B) * P(day 4, alive)
= P(find dog | in B) * P(in B) * P(day 4, alive) * P(day 3, alive) * P(day 2, alive) * P(day 1, alive)
= 0.15 * 0.6 * (2/5) * (2/4) * (2/3) * (2/2)
= 0.09 * 8/60
= 0.09 * 2/15

The solution gives the following reasoning:

P(find dog, in B, day 4, alive) = ...some magic...
= P(day 4, alive) * P(day 3, alive) * P(day 2, alive)
= (2/5) * (2/4) * (2/3)
= 2/15

I do not know how that is the accepted answer when they completely neglect the probability of P(find dog | in B) * P(in B) from the answer. The given answer is only the probability of the dog being alive on day 4. Intuitively, this feels lacking. Please help!

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  • $\begingroup$ You're wasting our time by posting two questions with the same mistakes without linking them: math.stackexchange.com/questions/3530180. $\endgroup$ – joriki Feb 1 at 11:26
  • $\begingroup$ @joriki, I am not wasting your time. That question is for e). This question is for f). $\endgroup$ – muxo Feb 1 at 11:32
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You’re misinterpreting the problem in (at least) two places.

First, the probability $\frac N{N+2}$ is not, as you seem to have assumed in your first calculation, the probability that the dog is dead on the $N$-th evening, but the probability that the dog dies on the $N$-th evening if it was alive until then.

Second, question f) does not ask for the probability to find the dog alive in $B$ on day $4$. It asks for the probability of finding the dog alive, given that it is found in $B$ on day $4$.

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  • $\begingroup$ I take $\frac{n}{n+2}$ as it means. I only use it to calculate the probability of the dog being alive on during the daytime of day n, which is equivalent to it being alive on the evening of day n - 1. $\endgroup$ – muxo Feb 1 at 10:33
  • $\begingroup$ As for question f) are you saying they are just asking for the probability of the dog being alive on day 4, regardless of the rescue attempt, the search location and its outcome? $\endgroup$ – muxo Feb 1 at 11:02
  • $\begingroup$ @retu: The error in the first calculation isn't related to the difference between days and evenings; I think you got that part right. The error is that you use $n/(n+2)$ as a probability of being dead rather than a probability of dying. The correct probability of being alive on day $3$ is $\left(1-\frac13\right)\left(1-\frac24\right)=\frac13$, whereas your formula yields $\frac24=\frac12$. $\endgroup$ – joriki Feb 1 at 11:11
  • $\begingroup$ @retu: As for question f): Effectively yes. Since the information that the dog was found on day $4$ is given, the question whether the dog was found alive is equivalent to the question whether the dog was alive on day $4$, and that's independent of rescue attempts, locations and outcomes. $\endgroup$ – joriki Feb 1 at 11:14
  • $\begingroup$ I have edited my question to clarify things. Using your method, the probability of being alive on day 3 is $(1-\frac{1}{3})(1-\frac{2}{4}) = (\frac{2}{3})(\frac{2}{4}) = \frac{1}{3}$. Using my method, it is $(\frac{2}{3})(\frac{2}{4}) = (\frac{1}{3})$. Would you agree? $\endgroup$ – muxo Feb 1 at 11:42

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