Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2979 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{n+1}$
$\mathrm{P}\left(\mathrm{X}_1^{\prime} \cap \mathrm{X}_2^{\prime} \cap \mathrm{X}_3^{\prime} \ldots \cap \mathrm{X}_{\mathrm{n}}^{\prime}\right)$
$\begin{aligned} & =P\left(X_1^{\prime}\right) P\left(X_2^{\prime}\right) P\left(X_3^{\prime}\right) \ldots P\left(X_n^{\prime}\right) \\ & =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots .\left(1-\frac{1}{n}\right)\left(1-\frac{1}{n+1}\right) \\ & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \ldots \times \frac{n-1}{n} \times \frac{n}{n+1}=\frac{1}{n+1}\end{aligned}$
$\begin{aligned} & =P\left(X_1^{\prime}\right) P\left(X_2^{\prime}\right) P\left(X_3^{\prime}\right) \ldots P\left(X_n^{\prime}\right) \\ & =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots .\left(1-\frac{1}{n}\right)\left(1-\frac{1}{n+1}\right) \\ & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \ldots \times \frac{n-1}{n} \times \frac{n}{n+1}=\frac{1}{n+1}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.