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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)=(1/2)*0.25+(1/2)*0.15;

but the answer given is=(0.5*0.4*0.25)+(0.5*0.6*0.15);

what is wrong with my logic?

3 Answers

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Using Bayes theorem 0.5*0.4*0.25/0.5*0.4*0.25+0.5*0.6*0.15=0.52631
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I hope this clarifies the doubt.

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According to me the answer should be (0.5*0.4*0.25)+(0.5*0.6*0.15) because 

On first day Oscar will search and get his dog only when his dog is lost at forest A and finds him or when his dog is lost at forest B and he finds him. These are with condition he flipping the coin.

Hence (0.5*0.4*0.25)+(0.5*0.6*0.15)  =  0.095

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