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The probability of catching fish in one hour is 0.64? What is the probability of catching fish in half an hour? Regular solution follows the principle for an opposite event: a probability of NOT catching fish in one hour is 1-0.64=0.36. Probability of not catching fish in an hour is not catching in first half of an hour multiplied by not catching in second half of an hour p' x p' = 0.36. Therefore p' = 0.6. Using opposite event again 1 - 0.6 = 0.4 is the probability of catching fish in half an hour.

But how to define the probability in terms only of successful event? Which formula should I use to calculate p directly? Is it possible at all?

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  • $\begingroup$ What you ask is much harder than this methodology since number of fish could go infinite. $\endgroup$
    – gunes
    Commented Jan 5, 2019 at 20:17

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I've come up with this solution (not sure whether it's correct though):

p(c) = 0.64 - probability of catching fish in one hour;

p(n) = 1 - p(c) = 0.36 - probability of not catching fish in one hour;

p(n) = p(hn) * p(hn) => p(hn) = 0.6 - probability of not catching fish in half an hour

p(c) = p(hc1)*p(hn1) + p(hc2)p(hn2) - total probability is the sum of each half an hour;

p(hn1) = 1 - we definitely not catching a fish because we have to catch it (that's the premise of our formula);

p(hc1) = p(hc2) - because probability of catching is equal across the time;

Therefore we have p(c) = p(hc1)(1 + p(hn2)) = 0.64 => p(hc1) = 0.64 / 1.6 = 0.4

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