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How do you calculate the probability of simultaneous events? As in, given four simultaneous events each with a 10% probability, what are the odds that ONE of them occurs? Obviously it isn't 40%, because...well, if you have ten events that probability clearly isn't 100%!

(Nope, not homework! Video games, probability of elemental effects from a given spell)

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  • $\begingroup$ Let A1, A2, A3, A4 be the events. Let P(A) be the probability of the event. So P(A1) = P(A2) = P(A3) = P(A4) = 0.1. You are searching for the probability of the union. P(A1 union A2 union A3 union A4) = P(A1) + P(A2) + P(A3) + P(A4) - P(A1)P(A2) - P(A1)P(A3) - P(A1)P(A4) - P(A2)P(A3) - P(A2)P(A4) - P(A3)P(A4) + P(A1)P(A2)P(A3) + P(A1)P(A2)P(A4) + P(A1)P(A3)P(A4) + P(A2)P(A3)P(A4) - P(A1)P(A2)P(A3)P(A4). Do the calculation and you'll have the exact probability. Hope this helps. $\endgroup$
    – user49557
    Apr 29, 2013 at 1:16
  • $\begingroup$ You are my hero. I have no grasp on binomials, so a step by step process is exactly what I needed. Put that in an answer so I can vote and accept it. ;) $\endgroup$ Apr 29, 2013 at 1:20
  • $\begingroup$ Nevermind, I'm new in the math section so I can't vote up. I can accept, though. $\endgroup$ Apr 29, 2013 at 1:25

4 Answers 4

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Whenever you are asked to find probability of at least ONE event of occurring, it is easier to find the probability of none of them to occur and remove it from $1$. In this case, $$1-(1-P(A))(1-P(B))(1-P(C))(1-P(D))=1-0.9\cdot0.9\cdot0.9\cdot0.9=0.3439$$

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Even though I lack sufficient background in probability (at time of writing), on request by the author, I'm following up on my comment on the question.

Let $A, B, C, D$ be the events. Let $P(X)$ be the probability of the event $X$. Then $P(A) = P(B) = P(C) = P(D) = 0.1$.

We are searching for the union of events:

$P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D)- P(A)P(B) - P(A)P(C) - P(A)P(D) - P(B)P(C) - P(B)P(D) - P(C)P(D) + P(A)P(B)P(C) + P(A)P(B)P(D) + P(A)P(C)P(D) + P(B)P(C)P(D) - P(A)P(B)P(C)P(D)$

Analogy:

Imagine the events as sets which can intersect one-another, and the probability of the event as the cardinality number of the corresponding set over the sum of cardinality numbers of all sets. Then for two sets, the cardinality number of the union is the sum of the cardinality numbers of the sets minus the cardinality number of their intersection so that we do not add those elements twice. For three sets, we sum up the cardinality numbers of the sets, we subtract the cardinality numbers of each intersection of two sets and then we add back the cardinality number of the intersection of all three sets, and so on. If you want to know more regarding the generalization of this principle, study the basics of combinatorics.

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  • $\begingroup$ I guess this would solve the problem for "at least" one of them occur, while it's not mentioned in the question. The OP has written "ONE" of them will occur, which means exactly one of them. $\endgroup$
    – aderchox
    Jan 10, 2021 at 15:10
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If these events are statisticallly independent, the number of them that occur is governed by a binomial distribution.

http://en.wikipedia.org/wiki/Binomial_distribution

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I just want to emphasize that the problem has NOT asked (at least not explicitly) for the probability of the occurrence of "at least" one of the events. What the question implies is the occurrence of "exactly" one of them, and in that case, as @ncmathsadist has said, it would be a Bernoulli experiment, i.e., an experiment with a sequence of independent trials, each only having "two" possible outcomes, e.g., occured or not-occured.

To calculate the probability of k successes(e.g., occured) in n trials in a Bernoulli experiment we would use this formula famously known as the binomial distribution: ${n \choose k} * p^k * (1-p)^{n-k}$ where p is the probability of the success of each single trial. (Note that also the logic behind this formula is fairly simple, it is summing up all the probabilities for k successfull trials among all the n independent trials.)

Now in this problem, $k$ is $1$, $n$ is $4$ and $p$ is $0.1$, so the final answer would be very easily achievable, it will give: $29\%$.

Also if what the OP really needs is the $p(at \ least \ one \ occurrence)$, it would be $1-p(no \ occurrences)$, which again contains a Bernoulli experiment, and can be calculated using the binomial distribution, which would give: $1 - {4 \choose 0}*0.1^0*0.9^4 = 1 - 0.9^4 = 34\%$ as @Vasil suggested.

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