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If $E_1, E_2 \ldots, E_n$ are an independent events such that $P\left(E_r\right)=\frac{1}{1+r},(r=1,2, \ldots, n)$, then the probability that atleast one of $E_1, E_2, \ldots, E_n$ happens is
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Verified Answer
The correct answer is:
$\frac{n}{n+1}$
We have,
$P\left(E_r\right)=\frac{1}{1+r}$
$\therefore P\left(\bar{E}_r\right)=1-P\left(E_r\right)=1-\frac{1}{1+r}=\frac{r}{1+r}$
$\therefore$ Required probability $=P\left(E_1 \cup E_2 \cup E_3 \ldots \cup E_n\right)$
$=1-P\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3 \ldots \cap \bar{E}_n\right)$
$=1-P\left(\bar{E}_1\right) P\left(\bar{E}_2\right) P\left(\bar{E}_3\right) \ldots P\left(\bar{E}_n\right)$
$=1-\left(\frac{1}{1+1}\right)\left(\frac{2}{1+2}\right)\left(\frac{3}{1+3}\right) \cdots\left(\frac{n}{1+n}\right)$
$=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{1+n}=1-\frac{1}{1+n}=\frac{n}{n+1}$
$P\left(E_r\right)=\frac{1}{1+r}$
$\therefore P\left(\bar{E}_r\right)=1-P\left(E_r\right)=1-\frac{1}{1+r}=\frac{r}{1+r}$
$\therefore$ Required probability $=P\left(E_1 \cup E_2 \cup E_3 \ldots \cup E_n\right)$
$=1-P\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3 \ldots \cap \bar{E}_n\right)$
$=1-P\left(\bar{E}_1\right) P\left(\bar{E}_2\right) P\left(\bar{E}_3\right) \ldots P\left(\bar{E}_n\right)$
$=1-\left(\frac{1}{1+1}\right)\left(\frac{2}{1+2}\right)\left(\frac{3}{1+3}\right) \cdots\left(\frac{n}{1+n}\right)$
$=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{1+n}=1-\frac{1}{1+n}=\frac{n}{n+1}$
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