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I can't for the life of me figure this problem out:

"There is a 0.01% of chance that your area might be hit by an earthquake. You heard that if an earth quake is coming, 90% of times that immediately prior to the earthquake, animals such as frogs will leave their shelters and appear on the street. You also know that, there is 1% of chance for the frogs to appear on the street when there is no earthquake. One day you see a very scary scene on the street: hundreds of thousands of frogs appear everywhere. What's the probability that your area is going to have an earthquake?"

Let e = earthquake, f = frogs.

From the description, I've gathered:

Facts: p(e) = .0001, p(f | e) = .9, p(f ^ -e) = .01.

Goal: find p(e | f)

From those facts (since the summation of probability distribution over a random variable should be “1”):

p(-e) = .9999, p(f | -e) = 0.1

I know p(e | f) = ((p(e)*p(f | e)) / p(f)). We have p(e) and p(f | e), so we just need to find p(f). We have p(f ^ -e) and p(f | -e), which should help us find p(f). However, I've tried writing out all permutations of Bayes' theorem, and I can't seem to figure out how to solve for p(f). I think I must be missing something simple.

Can someone please point out what I'm missing?

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    $\begingroup$ "There is a 1% chance for frogs to appear on the street when there is no earthquake" What does that mean? Is there a 99% chance they'll appear when there is an earthquake? No, that's 90%. $\endgroup$ – Kaynex Oct 27 '16 at 21:53
  • $\begingroup$ Hint : $P(F)=P(F|E)\cdot P(E)+P(F|-E)\cdot P(-E)$ (law of total probability) $\endgroup$ – Peter Oct 27 '16 at 21:53
  • $\begingroup$ Specifically, the line Kaynex points out is $p(f\mid \neg e)$. $\endgroup$ – Graham Kemp Oct 27 '16 at 21:57
  • $\begingroup$ So what Kaynex is saying is p(f | ¬e) is 0.9? And Peter, are you hinting that I shouldn't even be using Bayes' theorem for this? $\endgroup$ – Archer Oct 27 '16 at 22:04
  • $\begingroup$ @Archer No, you need Bayes' as well. Just calculate $P(E\cap F)$ and apply it. $\endgroup$ – Peter Oct 27 '16 at 22:51
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You have

  • $p(e)=0.0001$ as "There is a 0.01% of chance that your area might be hit by an earthquake."
    • So $p(\neg e) = 0.9999$ obvs.
  • $p(f\mid e)=0.90$ as "You heard that if an earth quake is coming, 90% of times that immediately prior to the earthquake, animals such as frogs will leave their shelters and appear on the street."
  • $p(f\mid \neg e) = 0.01$ as "You also know that, there is 1% of chance for the frogs to appear on the street when there is no earthquake."
    • Note: That is a conditional, not a conjunction.

Then by Bayes' Rule and the Law of Total Probability: $$p(e\mid f) = \dfrac{p(e)~p(f\mid e)}{p(e)~p(f\mid e)+p(\neg e)~p(f\mid\neg e)}$$

The rest is just substitution and calculation.


Remark: Don't be surprised by the result.   The probability of earthquake when frogs are on the street is quite low.   This is because earthquakes have such low incidence rate that the rate of false positives is still high enough for them to swamp out the much rarer true positives.

People, however, will always reminisce about "the day the danged frogs warned us about the earthquake" and forget about those other froggy days.

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  • $\begingroup$ That makes sense, thank you very much Graham! $\endgroup$ – Archer Oct 28 '16 at 4:44

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