Imagine I have two events A and B. On a day only one event can occur and if A occurs on the first day then A keeps on occurring and if B occurs then only B occurs from now on. Also P(A) = 0.05 and P(B) = 0.95. Also let there be another event C that occurs where, P(C | A) = 0.8 and P(C | B) = 0.1. How do I calculate the probability of C happening on the first, second and third days?
Q. Bayes Problem 2
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bayes-theorem
1 Answers
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Since $A$ and $B$ are mutually exclusive and $P(A)+P(B) = 1,$ we can write $$P(3C) = P(3C|A)P(A) + P(3C|B)P(B) $$ where $3C$ is the the event that $C$ occurs on the first three days. We can compute these as $$ P(3C|A) = P(C|A)^3 = \frac{512}{1000}\\P(3C|B) = P(C|B)^3 = \frac{1}{1000} $$ (You don't say much about the independence assumptions for $C$ between the days, but I presume it's independent conditional on whether $A$ or $B$ happened the first day.)
So we have $$ P(3C) = \frac{512}{1000}\frac{5}{100} + \frac{1}{1000}\frac{95}{100}= 0.02655$$