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I am trying to allocate probability to three mutually exclusive events. Two of the events depend on the first NOT being true. At least one of the events must occur, so the probabilities of all three must sum to 1.

Event 1 = P[E1]

Event 2 = P[E2] given Event 1 has NOT occurred, so would it be P[2]*1-P[1]?

Event 3 = Probability that neither Event 1 nor Event 2 have occurred. Would this just be 1-P[1]-P[2]? Why does (1-P2)*(1-P2) not work mathematically... it can end up being greater than 100% if.

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  • $\begingroup$ Given that $E_1$ has not occurred, the probability that $E_2$ did occur would be expected to increase. Imagine, for instance, that you are throwing a fair die and that $E_1$ is the event "$1,2,3$ come up", $E_2$ is the event "$4,5$ come up", and $E_3$ is the event "$6$ comes up". Then, given that $E_1$ does not occur, the probability that $E_2$ did occur is clearly $\frac 23$. $\endgroup$ – lulu Apr 24 '18 at 23:22
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Things to remember:

$Pr(\emptyset)=0$, $Pr(\Omega)=1$ where $\Omega$ is the sample space and $0\leq Pr(A)\leq 1$ for all events $A$.

Inclusion Exclusion (two event case): $Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B)$

Definition of conditional probability: $Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$. Rearranged this says that $Pr(A\cap B)=Pr(B)\cdot Pr(A\mid B)$. Independent Events are those special pairs of events with the special property that $Pr(A\cap B)=Pr(A)Pr(B)$ but this is explicitly not true in general and in particular is never true for dependent events.

Mutual Exclusive events: $A$ and $B$ are said to be mutually exclusive iff $A\cap B=\emptyset$. This of course then implies that if $A$ and $B$ are mutually exclusive then $Pr(A\cap B)=0$ and further that $Pr(A\cup B)=Pr(A)+Pr(B)$.

Total probability: $Pr(A)+Pr(A^c)=1$ and similarly $Pr(A\cap B)+Pr(A\cap B^c)=Pr(A)$


So, the probability that $E_1$ occurred is notated as $Pr(E_1)$. There is nothing more to calculate for the first part of the question.

The probability that $E_2$ occurs given that $E_1$ has not occurred:

$Pr(E_2\mid E_1^c)=\dfrac{Pr(E_2\cap E_1^c)}{Pr(E_1^c)}=\dfrac{Pr(E_2)-Pr(E_2\cap E_1)}{1-Pr(E_1)}=\dfrac{Pr(E_2)}{1-Pr(E_1)}$

As for the third part of the question, the probability that neither $E_1$ nor $E_2$ have occurred is simply $Pr(E_3)$, alternatively written as $1-Pr(E_1)-Pr(E_2)$, since you stated that $E_1,E_2,E_3$ are mutually exclusive and exhaustive. The probability that $E_3$ occurs given that $E_1$ and $E_2$ does not occur can be expanded as above and simplifies simply to $Pr(E_3\mid E_1^c\cap E_2^c)=1$

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