If A_i (i=1, 2, 3, ldots, n) are n independen | vedclass

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(D) The probability that event $A_i$ does not occur is given by $P(\bar{A}_i) = 1 - P(A_i)$.Given $P(A_i) = \frac{1}{1+i}$, we have $P(\bar{A}_i) = 1

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

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(D) The probability that event $A_i$ does not occur is given by $P(\bar{A}_i) = 1 - P(A_i)$.Given $P(A_i) = \frac{1}{1+i}$, we have $P(\bar{A}_i) = 1 - \frac{1}{1+i} = \frac{1+i-1}{1+i} = \frac{i}{1+i}$.Since $A_i$ are independent events, the probability that none of $A_i$ occurs is the product of the probabilities of their complements:$P(	ext{none occurs}) = P(\bar{A}_1) 	imes P(\bar{A}_2) 	imes \ldots 	imes P(\bar{A}_n)$.Substituting the values:$P(	ext{none occurs}) = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} 	imes \ldots 	imes \frac{n}{n+1}$.Observing the product, the terms cancel out in a telescoping manner:$P(	ext{none occurs}) = \frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4} 	imes \ldots 	imes \frac{n}{n+1} = \frac{1}{n+1}$.

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

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