For independent events $A_1, A_2, A_3, ..., A_n$ if $P(A_i) = \frac{1}{i+1}$, $i = 1, 2, 3, ..., n$, then the probability that none of the events occur is:

Events $A_1, A_2, A_3, ..., A_n$ are independent with $P(A_i) = \frac{1}{i+1}$.

This gives:

• $P(A_1) = \frac{1}{2}$
• $P(A_2) = \frac{1}{3}$
• $P(A_3) = \frac{1}{4}$
• And so on...

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"None of the events occur" means $A_1$ doesn't happen AND $A_2$ doesn't happen AND $A_3$ doesn't happen, etc.

In probability notation: $P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \cap ... \cap \overline{A_n})$

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Since the events are independent:

$P(\text{none occur}) = P(\overline{A_1}) \times P(\overline{A_2}) \times P(\overline{A_3}) \times ... \times P(\overline{A_n})$

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For each event, using the complement rule:

$P(\overline{A_i}) = 1 - P(A_i)$

$P(\overline{A_i}) = 1 - \frac{1}{i+1}$

$P(\overline{A_i}) = \frac{i+1-1}{i+1}$

$P(\overline{A_i}) = \frac{i}{i+1}$

This gives:

• $P(\overline{A_1}) = \frac{1}{2}$
• $P(\overline{A_2}) = \frac{2}{3}$
• $P(\overline{A_3}) = \frac{3}{4}$
• $P(\overline{A_n}) = \frac{n}{n+1}$

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$P(\text{none occur}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times ... \times \frac{n}{n+1}$

$= \frac{1 \times 2 \times 3 \times 4 \times ... \times n}{2 \times 3 \times 4 \times 5 \times ... \times (n+1)}$

The 2 in the numerator cancels with 2 in the denominator, the 3 cancels with 3, the 4 with 4, and so on.

After cancellation, only 1 remains in the numerator and only $(n+1)$ remains in the denominator.

$= \frac{1}{n+1}$

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Therefore, the probability that none of the events occur is $\frac{1}{n+1}$.