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

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!

• Shouldn't the probabilities sum up to one? Or is it possible that I don't collect the money? Commented May 17, 2019 at 16:35
• @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. Commented May 17, 2019 at 16:41

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.

• 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! Commented May 17, 2019 at 16:47
• You're welcome! 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)?

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

• 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. Commented May 17, 2019 at 16:53