How does the probability of events change if an event does not occur

probabilityprobability distributions

Suppose that someone tells me I will collect $\$100$ dollars within some time interval. Those time intervals are 1 to 7 days, 8 to 30 days and eventually after 30 days.

Let $A$ be the event I collect the $\$100$ dollars in 1 to 7 days, $B$ be the event I collect the $\$100$ dollars in 8 to 30 days and $C$ be the event that I collect the $\$100$ dollars eventually after 30 days.

Let $P(A)=0.40, P(B) = 0.50$ and $P(C)=0.10$.

The events are mutually exclusive, so once I have collected the $\$100$ dollars I can not collected it at another time and the $\$100$ dollars can also not be broken up into different intervals.

The events are also temporal so $B$ can only happen if $A$ does not happen and $C$ can only happen if $B$ does not happen, so there is a dependency.

My question is how do these probabilities change after an event has passed?

  1. If $A$ does not happen how does that affect the $P(B)$ and the $P(C)$?
  2. Going further if both $A$ and $B$ do not happen then how does that affect $P(C)$?

Also, I don't know if this is a textbook case, but this is not a homework question. It is an actual problem that I am wondering about but my probability theory is not strong enough to solve on my own.

Thank you for any help and comments!

Best Answer

What you're looking for is known as conditional probability. In general, the probability of $A$ given $B$, written $P(A|B)$, is given by $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ Intuitively, you can interpret this as once $B$ occurs, we are in the world of $B$, so our new overall probability is $P(B)$. The probability of $A$ in this new world is no longer $P(A)$, but $P(A\cap B)$, since $B$ has already occurred.

In this case, your questions can be solved as follows:

  1. $$P(B|A^{c})=\frac{P(B\cap A^{c})}{P(A^{c})}=\frac{P(B)}{P(A^{c})}=\frac{.5}{.6}=\frac{5}{6}\approx .83$$ and similarly $$P(C|A^{c})=\frac{P(C\cap A^{c})}{P(A^{c})}=\frac{P(C)}{P(A^{c})}=\frac{.095}{.6}\approx .158$$

  2. $$P(C|A^{c}\cap B^{c})=\frac{P(C\cap A^{c}\cap B^{c})}{P(A^{c}\cap B^{c})}=\frac{P(C)}{P(A^{c}\cap B^{c})}=\frac{P(C)}{P(A^{c})\cdot P(B^{c}|A^{c})}=\frac{.095}{.6(1/6)}\approx .95$$

Note that $P(B\cap A^{c})=P(B)$, $P(C\cap A^{c})=P(C)$, and $P(C\cap A^{c}\cap B^{c})=P(C)$ all follow from mutual exclusivity.

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