If E_1, E_2, ldots, E_n are independent event | vedclass

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(C) Given that $E_1, E_2, \ldots, E_n$ are independent events with $P(E_r) = \frac{1}{1+r}$.First, we find the probability of the complement event $\b

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given that $E_1, E_2, \ldots, E_n$ are independent events with $P(E_r) = \frac{1}{1+r}$.First, we find the probability of the complement event $\bar{E}_r$ for each $r$:$P(\bar{E}_r) = 1 - P(E_r) = 1 - \frac{1}{1+r} = \frac{r}{1+r}$.The probability that at least one of the events occurs is given by $1 - P(	ext{none of the events occur})$.Since the events are independent, the probability that none of the events occur is the product of the probabilities of their complements:$P(	ext{none}) = P(\bar{E}_1) 	imes P(\bar{E}_2) 	imes \cdots 	imes P(\bar{E}_n)$.Substituting the values:$P(	ext{none}) = \left(\frac{1}{2}ight) 	imes \left(\frac{2}{3}ight) 	imes \left(\frac{3}{4}ight) 	imes \cdots 	imes \left(\frac{n}{n+1}ight)$.This is a telescoping product where the numerator of each term cancels with the denominator of the previous term:$P(	ext{none}) = \frac{1}{n+1}$.Therefore, the probability that at least one event happens is:$1 - P(	ext{none}) = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

If $E_1, E_2, \ldots, E_n$ are independent events such that $P(E_r) = \frac{1}{1+r}$ for $r = 1, 2, \ldots, n$, then the probability that at least one of $E_1, E_2, \ldots, E_n$ happens is

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