TITLE: Probability to die in the first or second term?
QUESTION [0 upvotes]: Apologies for the morbid example.
If there's a 20% chance for a candidate to die in their first term (A) and a 32% chance in their second term (B|A?). What is the probability of them dying in either the first or second term?
From my understanding these are two mutually exclusive events, is that right? If not why?
When P(A) = 0.2 and P(B|A) = 0.32
If mutually exclusive then P(A OR B) = P(A) + P(B|A) = 0.52 = 52%
If not mutually exclusive then P(A OR B) = P(A) + P(B|A) - P(A) * P(B|A) = 0.456 = 45.6%
Which is it and why?
Also can a dependent event be mutually exclusive? If the occurrence of an event prevents the occurrence of another, that fits both definitions correct?

REPLY [0 votes]: You are correct that these events are mutually exclusive.  If an official dies in his/her first term, then he or she cannot die in his/her second term since he/she is already dead.
There are two "favorable" possibilities (an odd term in this case).  One is that the candidate dies in his/her first term.  The other is that he/she survives the first term but dies in his/her second term.  We are given that
\begin{align*}
\Pr(A) & = 0.2\\
\Pr(B \mid A') & = 0.32
\end{align*}
where $A'$ is the event that the official does not die in his/her first term, which occurs with probability $\Pr(A') = 1 - \Pr(A) = 0.8$.  Hence, the probability that the official dies while in office, assuming he or she serves a second term after surviving the first (something which is not normally guaranteed to occur), is
$$\Pr(A~\text{OR}~B) = \Pr(A) + \Pr(A')\Pr(B \mid A') = 0.2 + (0.8)(0.32) = 0.456$$
While you obtained that answer with your second method, your reasoning does not make sense.
Clearly, the occurrence of event $B$ depends on event $A$ not occurring.
{"set_name": "stack_exchange", "score": 0, "question_id": 3792170}