If $A_i(i=1,2,3, \ldots, n)$ are $n$ independent events with $P\left(A_i\right)=\frac{1}{1+i}$ for each $i$, then the probability that none of $A_i$ occurs is

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Aicharya · Accepted Answer

The required probability $=P\left(\bar{A}_1 \cap \bar{A}_2 \cap \ldots \cap \bar{A}_n\right)$ $=P\left(\bar{A}_1\right) P\left(\bar{A}_2\right) \ldots P\left(\bar{A}_n\right)$ $=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \cdot \frac{n}{n+1}$ $=\frac{1}{n+1}$

If $A_i(i=1,2,3, \ldots, n)$ are $n$ independent events with $P\left(A_i\right)=\frac{1}{1+i}$ for each $i$,…

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