1 votes 1 votes Let $A_1,A_2,A_3, \dots , A_n$ be $n$ independent events such that $P(A_i) = \frac{1}{i+1}$ for $i=1,2,3, \dots , n$. The probability that none of $A_1, A_2, A_3, \dots , A_n$ occurs is $\frac{n}{n+1}$ $\frac{1}{n+1}$ $\frac{n-1}{n+1}$ none of these Probability isi2017-dcg probability independent-events + – gatecse asked Sep 18, 2019 • recategorized Nov 15, 2019 by Lakshman Bhaiya gatecse 479 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Given $P(A{_{i}})= \frac{1}{i + 1}$ Since all events are independent, probability that none of them occurs is $(1-P(A{_{1}})).(1-P(A{_{2}})).(1-P(A{_{3}}))\ldots(1-P(A{_{n}}))$ $\qquad =(1-\frac{1}{2}).(1-\frac{1}{3}).(1-\frac{1}{4})\ldots (1-\frac{1}{n+1})$ $\qquad =(\frac{1}{2}).(\frac{2}{3}).(\frac{3}{4})\ldots (\frac{n-1}{n}).(\frac{n}{n+1})$ $\qquad =\frac{1}{n+1}$ Option (B) is correct Ashwani Kumar 2 answered Sep 20, 2019 • selected Sep 6, 2020 by Arjun Ashwani Kumar 2 comment Share Follow See all 2 Comments See all 2 2 Comments reply priyankabedekar commented Dec 15, 2019 reply Follow Share Isn't it 1/n+1 as n-1 also gets canceled. So option B OR n!/(n+1)! Which is 1/n+1. 2 votes 2 votes hiteshsp commented Jan 29, 2020 reply Follow Share Yes I'm getting 1/n+1. which is correct ? 0 votes 0 votes Please log in or register to add a comment.