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!

  • 1
    $\begingroup$ Shouldn't the probabilities sum up to one? Or is it possible that I don't collect the money? $\endgroup$
    – leonbloy
    Commented May 17, 2019 at 16:35
  • $\begingroup$ @leonbloy - yes, very good question thank you for asking. The probability may not fully sum to 1 because it is possible that the money may not ever be collected but that chance is very rare but I will actually change it to sum to 1 since that does not really ever happen. $\endgroup$
    – RDizzl3
    Commented May 17, 2019 at 16:41

2 Answers 2


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.

  • $\begingroup$ Thank you @pwerth - this is exactly what I was looking for! I wanted a mathematically clean way to update the probabilities after the event had occurred! Thanks! $\endgroup$
    – RDizzl3
    Commented May 17, 2019 at 16:47
  • $\begingroup$ You're welcome! $\endgroup$
    – pwerth
    Commented May 17, 2019 at 16:49

Such events that are mutually exclusive and exhaustive are commonly modelled as a single variable $X$ that can take four values $(A,B,C,D)$ or $(1,2,3,4)$ (I added the event that you don't receive the money). Then we have $P(X=A)=0.1$ etc.

Then you apply conditional probabilities

If A does not happen how does that affect the P(B) and the P(C)?

$$P(X=B| X \ne A) = \frac{P(X \ne A | X=B )P(X=B)}{P(X\ne A)}=\frac{1 \times 0.5}{1-0.4}$$

if both A and B do not happen then how does that affect P(C)?

$$P(X=C| X \ne A \wedge X\ne B) = \frac{P(X \ne A \wedge X\ne B | X=C )P(X=C)}{P(X \ne A \wedge X\ne B)}=\frac{1 \times 0.095}{0.0095 + 0.005}$$

  • $\begingroup$ thank you so much for the insight as well. I upvoted this answer. Also, thank you for keeping in the probability of the event possibly not occuring. Quick question, does ∧ mean and? I am not familiar with that notation. $\endgroup$
    – RDizzl3
    Commented May 17, 2019 at 16:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .