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We have the events $D,F$ and $G$. Their probabilities are $$P(D)=\cfrac{1}{6},\quad P(F)=\cfrac{1}{8} \;\;\mbox{ and }\;\; P(G)=\cfrac{1}{4}.$$ Find the probability that any of these events happens( It does not specify which in my book). My opinion: Wouldnt that be $1$?


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marked as duplicate by Did, rschwieb, Micah, Henry T. Horton, tomasz Mar 20 '13 at 21:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

how did you form your opinion? the answer can vary from 1/4 to 13/24 depending how the events are related... – Lost1 Mar 20 '13 at 19:58
Why ask the same question twice?… – Tony Mar 20 '13 at 20:03

If the events are mutually exclusive, which means that only one can happen, you can add the probabilities and get $\frac 16 + \frac 18 + \frac 14=\frac {13}{24}$

If the events are independent, so whether one happens has no effect on whether another one does, the chance of at least one happening is lower because some of the time more than one will happen. The chance of none of them happening is the product of the chance of each of them not happening because they all have to not happen. This is $\frac 56 \cdot \frac 78 \cdot \frac 34=\frac {105}{192}=\frac {35}{64}$. The chance that at least one happens is this subtracted from $1$, or $\frac {29}{64}$

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