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I am trying to solve the following problem and was wondering if someone can verify my answers.

Big Joe has lost an important document. There is a 70% probability it is at home, and a 30% chance it is in his office He has a 50% chance of finding it if it is at home and 80% chance of finding the document if it is in the office.

Questions:

  1. Where should he look first to maximise his opportunity of finding the document?
  2. What is the probability that the document is still lost after the search?
  3. Assume he did not find the document at the location in part 1. Where should he look next? What is the probability of not finding the document there?

Answers

  1. The probability of finding it at home $P(Home \cap Found) = 0.35$ while $P(Office \cap Found) = 0.24$.
  2. Probability that document is still lost $= 0.41 (P(Home \cap Not Found)+P(Office \cap Not Found)$)
  3. Using Bayes theorem, $P(Home \cap Not Found) = 0.6$. This is based on the following equation, $\frac {P(Not Found | Home).P(Home)}{ P(Not Found)}$ Based on the above, he should search his home again and the probability of not finding there will now be 0.3.

This question is probababyl quite trivial but I'm a novice so any help will be highly appreciated.

Thanks

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Your first answers is correct. In the second one, either the document is in the office, or at home and you didn't find it, i.e. simply, $P(\bar{H})+P(H)P(\bar{F}|H)=0.3+0.7 \times 0.5=0.65$. You pretend as if you also searched it in office, but you didn't. In the third one, we'll update our $P(H)$ based on this experiment. $$P(H|\text{Home-Search Fails})=\frac{P(\text{Home-Search Fails}|H)P(H)}{P(\text{Home-Search Fails})}=\frac{0.35}{0.65}=\frac{7}{13}$$ based on parts (a) and (b). Now, new $P(H)$ and $P(O)$ are $7/13$ and $6/13$. Do the same in (a).

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  • $\begingroup$ Thanks Gunes.It took me a while to understand what the denominator should be 0.65 but this now makes a lot of sense. $\endgroup$ – Electromagnet Jan 16 at 9:20

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