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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 $0.6$). On any given day, if the dog is in A 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, the conditional probability that he will find the dog that day is $0.15$. 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.

Q> If Oscar flips a fair coin to determine where to look on the first day and finds the dog on the first day, what is the probability that he looked in A?

According to me P(finding the dog) = $\dfrac{1}{2}\cdot 0.25 + \dfrac{1}{2}\cdot 0.15$;

but the answer given is $(0.5\cdot 0.4\cdot 0.25)+(0.5\cdot0.6\cdot0.15)$;

what is wrong with my logic?

F.A.
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Diya
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    You appear to be trying to compute the probability that Oscar finds the dog on the first day, which is not what was asked. Also, you ignore the fact that the dog is probably in $B$. – lulu Feb 16 '19 at 14:34
  • My logic goes this way->There is 1/2 probability that Oscar searches in A(Here already I am getting the place where I am searching, that is A).
    Then why should we multiply 0.4 with it in the expression 0.50.40.25?
    Why isn't it 0.5 * 0.25 valid?. Can you please tell me where I am going wrong.     – Diya Feb 16 '19 at 15:29
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    Because you need the dog to be in $A$ to have any chance. Look at the extreme case. Suppose the dog is certain to be in $B$. Then the probability that you find him by searching in $A$ is $0$. – lulu Feb 16 '19 at 15:31
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    For Oscar to find the dog in $A$, three separate things have to happen. First, he must look in $A$. Second, the dog has to be in $A$. Third, he has to find the dog, – lulu Feb 16 '19 at 15:32
  • Understood.Here I was assuming that the dog is already in A which is wrong and we need to multiply the probability of the dog being in A. Thank you.:) – Diya Feb 16 '19 at 16:41

1 Answers1

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Neither your answer nor the official answer resolves the actual question. Rather, both attempt to answer the question "what is the probability that Oscar finds the dog on the first day."

To answer that question, note that there are two ways he can find the dog on the first day and both of those require several things to happen at once. He might look in $A$, the dog might be in $A$, and he might actually find the dog in $A$, or he might look in $B$, the dog might be in $B$, and he might actually find the dog in $B$. This leads to the official result, namely $$.5\times .4\times .25+.5 \times .6\times .5$$

Note that your answer ignores the probability that the dog is (or is not) actually in the forest Oscar's coin selects.

To solve the given problem, we need to ask what portion of this is explained by Oscar having searched in $A$. Thus the answer to the given problem is $$\frac {.5\times .4\times .25}{.5\times .4\times .25+.5 \times .6\times .5}\approx .526$$

lulu
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