Suppose that I have $30,000$ cases, and each case is assigned with a probability. Let's call these cases $a_1, a_2,.. a_{30000}$ and the assigned probabilities $P_1, P_2,.. P_{30000}$ ( the sum of these probabilities is equal to $1$). Among all these cases, I'm just interested in the probabilities of two cases ($A_5$ and $A_9$). Is there any statistical method to transform this multi probability case to binary for just these two cases?
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asked Mar 9, 2022 at 16:33
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2$\begingroup$ $P_5$ and $P_9$ are the probabilities, aren't they..? If not, it's not clear what you mean. $\endgroup$– TimCommented Mar 9, 2022 at 16:40
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$\begingroup$ @Tim They are, but considering the the rest of probabilities (29998 remaining ones). Here, I mean I want to isolate the probabilites to just two cases ($P_5$ and $P_9$). So we can have $P_{5} = 1 - P_{9}$. $\endgroup$– user2991243Commented Mar 9, 2022 at 16:42
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$\begingroup$ What do you mean by "isolating" them? If that is the case, you cannot have $P_5 = 1 - P_9$ because it can be that neither of the events happens. Unless you mean that you want to condition on that either of the events happened? $\endgroup$– TimCommented Mar 9, 2022 at 16:59
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2$\begingroup$ Are you perhaps asking how to compute the conditional probability distribution of $(a_5,a_9),$ conditional on the event that one of them is observed? Or maybe you want to collapse the event $\{a_5,a_9\}$ into one outcome and the remaining 29,998 cases into another outcome? And what would be the purpose or statistical setting for doing either one? $\endgroup$– whuber ♦Commented Mar 9, 2022 at 17:25
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1 Answer
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Reading between the lines, you seem to be saying that you have $a_1,a_2,\dots,a_n$ mutually exclusive events, hence $\sum_{i=1}^n P_i = 1$. You seem to be asking how to calculate the conditional probability of observing one of the events (say $a_5$) given that either of them occurred. It is just
$$ \Pr(a_5 | a_5 \lor a_9) = \frac{\Pr(a_5 \lor [a_5 \land a_9])}{\Pr(a_5 \lor a_9)} = \frac{\Pr(a_5 )}{\Pr(a_5 \lor a_9)} = \frac{P_5}{P_5 + P_9} $$
by the basic rules of boolean algebra and definition of conditional probability. Other than conditioning, I cannot see how you could "isolate" them.
