31 votes 31 votes $A$ and $B$ are friends. They decide to meet between 1:00 pm and 2:00 pm on a given day. There is a condition that whoever arrives first will not wait for the other for more than $15$ minutes. The probability that they will meet on that day is $1/4$ $1/16$ $7/16$ $9/16$ Quantitative Aptitude gate2012-cy quantitative-aptitude probability + – Arjun asked Nov 15, 2014 edited Jun 2, 2019 by Lakshman Bhaiya Arjun 17.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Mark this formula for this kind of any question. x=minutes one could not wait for. n=60/x. probability they meet= (2n-1)/n^2 here x=60/15 =4. probabi;ity=8-1/16 =7/16 Sanjay Kumar 7 answered Jan 14, 2019 Sanjay Kumar 7 comment Share Follow See 1 comment See all 1 1 comment reply Pranavpurkar commented Aug 26, 2022 reply Follow Share Sanjay Kumar 7 here x=60/15 =4. it should be n=4. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes . MANSI_SOMANI answered Oct 22, 2022 MANSI_SOMANI comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes We can derive one formula also from graph For(n*n) (n^2-(n-1)^2) /n^2 from given ques n=4 (16-9)/16 7/16 Final answer akshat sharma answered Aug 24, 2017 akshat sharma comment Share Follow See 1 comment See all 1 1 comment reply MIRIYALA JEEVAN KUMA commented Nov 3, 2017 reply Follow Share Could you explain bit more please. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes C Gajanan Purud answered Sep 15, 2023 Gajanan Purud comment Share Follow See all 0 reply Please log in or register to add a comment.