Why the calculation does not take these probabilities into account?

Question

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!

Answer 1

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 $4$.