# Probability of complement events over time

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?

• What you ask is much harder than this methodology since number of fish could go infinite. – gunes Jan 5 at 20:17

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