What is the probability of multiple independent events occurring in a given amount of time? I thought about this problem the other day and my limited knowledge in basic probability theory was not enough to help me figure this one out. Say you have 3 probabilities:


You have a 50% chance of dying tomorrow due to ant abduction(P(A) = 0.5).
You have a 40% chance of dying tomorrow by being beaten by a baboon (P(B) = 0.4).
You have a 30% chance of dying tomorrow by being cannibalized by Canadians (P(C) = 0.3).


Then, what is the probability that you will die tomorrow, assuming the previous three reasons are the only ways you could die?

I thought of doing something like P(A or B or C), and since A,B,C are disjoint, then P(A or B or C) = P(A) + P(B) + P(C). This is clearly wrong, since the probabilities can't exceed 1. What would be an appropriate way of thinking about this?
The probability that you will not die tomorrow, is the probability all events will not happen:
$$\mathbb{P}(\text{you will not die})=(1-0.5)(1-0.4)(1-0.3)=\frac{21}{100}$$
So the probability that you will die tomorrow is:
$$\mathbb{P}(\text{you will die})=1-\mathbb{P}(\text{you will not die})=1-\frac{21}{100}=\frac{79}{100}$$
This problem may be easily solved using the common formula.
From the given situation it is obvious that any one of the happenings from three you have described is enough for your death.
You tried in good way. Just few more calculations was needed. The formula you used was almost correct. The correct formula is as follows:
$$
\begin{split}
P(A\cup B\cup C)&amp;=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A) + P(A \cap B \cap C)\\
&amp;= (0.5) + (0.4) + (0.3) - (0.5\cdot 0.4) + (0.4\cdot 0.3) + (0.3\cdot 0.5) + (0.5\cdot0.4\cdot0.3)\\
&amp;= 0.79
\end{split}
$$
See more here: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle