Q. Bayes' rule problem

Question

Imagine I have two events $A$ and $B$. On a day only one event can occur and $P(A) = 0.05$ 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?

I am assuming Bayes' rule would be good for this but I can't figure out the exact Formula for this.

What if either of $A$ or $B$ can only occur in all 3 days? Does that change the probability? Meaning if A occurs on day 1 it also occurs on all 3 days. Just a curious question.

Answer 1

You have been informed that on any day ($k\in\{1,2,3,....\}$) that: $~ \mathsf P(A_k)=0.05~,$ $~\mathsf P(B_k)=0.95~,$ $~ \mathsf P(C_k\mid A_k)=0.80~$, $~ \mathsf P(C_k\mid B)=0.10~$, and that $A_k,B_k$ are mutually exclusive (disjoint), and all events are independent of events on other days.

Then by the Law of Total Probability

$$\mathsf P(C_k)~=~\mathsf P(C_k\mid A_k)~\mathsf P(A_k)+\mathsf P(C_k\mid B_k)~\mathsf P(B_k)$$

Then because of daily independence:

$$\mathsf P(C_1,C_2,C_3) = \mathsf P(C_1) \mathsf P(C_2) \mathsf P(C_3)$$

Answer 2

You can use the Law of Total Probability. $$P(C)=P(A)\cdot P(C|A)+P(B)\cdot P(C|B)$$

Thus, for your question, for one day, $P(C)=0.05\cdot0.8+0.95\cdot0.1=0.135$.

Assuming days are independent, $P(C\text{ happening on first, second, third days})=0.135^3\approx0.00246$.