Conditional Probability(MIT Assignment) 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 nd 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 nd 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 ips a fair coin to determine where to look on the rst day and nds the dog on the rst 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?

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 $$\frac {.5\times .4\times .25}{.5\times .4\times .25+.5 \times .6\times .5}\approx .526$$