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I am trying to compute the probability of none the events occurs where the probability for each event is $Pr[A_i]=\frac{1}{n-1}$ for all i and these events are independent.

What is the $\prod_{i=3}^{n} \frac{1}{n-1}$ when n >= 3

I know that the Pr(none event occur) = 1 - Pr(at least one occur)

= 1 - $\prod_{i=3}^{n} \frac{1}{n-1}$

I want to proof that the probability that none of them occur is ≥1/8

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    You probably mean $\frac 1{i-1}$ instead of $\frac 1{n-1}$ ... right ? – 2017-02-15
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    @Adren Pr[Ai] = 1/n-1 for all i – 2017-02-15
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    Well $\prod_{i=3}^n A = \overbrace{A \cdot A \cdots A}^{n-2\text{ factors}} = A^{n-2}$ whenever $A$ does not depend on $i$. – 2017-02-15
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    The probability that none of the events occur can be easily computed straightforward... – 2017-02-15
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    What is the probability that an event does not occur? – 2017-02-15
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    @max it is 1 - 1/n-1. should I sum this value to get 1/8 ? – 2017-02-15
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    Correct, for one event it is n-2/n-1. Now since the events are independent, you can multiply the probabilities to obtain the probability that none of the events occur. – 2017-02-15

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  • $\prod_{1}^{n} \frac{1}{n-1} = (\frac{1}{n-1})^{n}$
  • $\Pr($none event occur$) = 1-(\frac{1}{n-1})^{n}$

For any $n\geq 3$, we have

$log(\frac{8}{7}) < n.log(n-1) \Rightarrow -log(\frac{8}{7}) > -n.log(n-1) \Rightarrow log(\frac{8}{7})^{-1} > n.log(n-1)^{-1} \Rightarrow log(\frac{7}{8}) > n.log(\frac{1}{n-1}) \Rightarrow \frac{7}{8} > (\frac{1}{n-1})^{n} \Rightarrow 1-\frac{1}{8}>(\frac{1}{n-1})^{n} \Rightarrow 1-(\frac{1}{n-1})^{n} >\frac{1}{8} \Rightarrow \Pr(none\text{ }event\text{ }occur) > \frac{1}{8}$

(All logarithms are in base 2)