If $\mathrm{X}_1, \mathrm{X}_2, \ldots \mathrm{X}_{\mathrm{n}}$ are $\mathrm{n}$ independent events such…

If $\mathrm{X}_1, \mathrm{X}_2, \ldots \mathrm{X}_{\mathrm{n}}$ are $\mathrm{n}$ independent events such that $\mathrm{P}\left(\mathrm{X}_{\mathrm{r}}\right)$ $=\frac{1}{r+1}, r=1,2, \ldots, n$, then the probability that none of the $n$ events occur is
  1. $\frac{1}{n}$
  2. $\frac{1}{n+1}$
  3. $\frac{\mathrm{n}}{\mathrm{n}+1}$
  4. $\frac{n+1}{n+2}$

Solution

$\mathrm{P}\left(\mathrm{X}_1^{\prime} \cap \mathrm{X}_2^{\prime} \cap \mathrm{X}_3^{\prime} \ldots \cap \mathrm{X}_{\mathrm{n}}^{\prime}\right)$ $\begin{aligned} & =P\left(X_1^{\prime}\right) P\left(X_2^{\prime}\right) P\left(X_3^{\prime}\right) \ldots P\left(X_n^{\prime}\right) \\ & =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots .\left(1-\frac{1}{n}\right)\left(1-\frac{1}{n+1}\right) \\ & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \ldots \times \frac{n-1}{n} \times \frac{n}{n+1}=\frac{1}{n+1}\end{aligned}$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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