7 votes 7 votes A fair die is tossed two times. the probability that 2nd toss results in value greater than first toss is ? Probability gate-ec probability expectation + – pC asked Jan 3, 2016 • retagged Jun 26, 2019 by Cristine pC 2.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 10 votes 10 votes total outcomes = 36 expected outcomes={(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}=15 hence Probability=15/36=5/12 Lakshya Mission answered Jan 3, 2016 • selected Jan 3, 2016 by Pooja Palod Lakshya Mission comment Share Follow See all 4 Comments See all 4 4 Comments reply pC commented Jan 3, 2016 reply Follow Share Pls explain me how are you writing this sample space 0 votes 0 votes Lakshya Mission commented Jan 3, 2016 reply Follow Share as it is mentioned in question : 2d toss results in value greater than first so if you assume that first toss results in outcome as : 2 now in case of tossing the coin 2nd time , allowed outcomes will be only : (2,3),(2,4),(2,5),(2,6) here we cannot include (2,1) and (2,2) as 1<2 and 2=2 but we are only allowed to have a greater value in case of 2nd toss! hope u will get it! 0 votes 0 votes pC commented Jan 3, 2016 reply Follow Share Ok. Thx :) I was confused by ordered pair (x,y) x<y x belongs to first throw and y belongs to sexond throw . , now i get it 0 votes 0 votes Lakshya Mission commented Jan 3, 2016 reply Follow Share welcome! :) 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Answer is {n(n-1)/2}/{n^2} = (n-1)/(2*n) Here n = 6 so answer is 5/12. Chhotu answered Oct 12, 2017 Chhotu comment Share Follow See all 3 Comments See all 3 3 Comments reply Puja Mishra commented Jan 25, 2018 reply Follow Share {n(n-1)/2} .... for what ?? i am nt getting it ... r u considering (5,6) as 1 ... (4,5) ,(4,6) as 2 ... ?? 0 votes 0 votes Anand. commented May 16, 2018 reply Follow Share given answer is absolutely wrong! 0 votes 0 votes chollu commented Dec 23, 2019 reply Follow Share This is how the formula is derived, On the first die is rolled getting an outcome is 1/6 On the second roll we need a value greater than the value appeared on the first roll, ie if the first outcome was 1, we can have 5 values greater than 1, so this probability is 5/6. if the first outcome was 2, we can have 4 values greater than 2 in second roll, so this probability is 4/6 and so on we get 3/6 for 3, 2/6 for 4, 1/6 for 5. So the required probability is given by, [1/6]*[(5/6)+(4/6)+(3/6)+(2/6)+(1/6)] = [1/(6^2)]*[1+2+3+4+5] which is equivalent to above expression. The answer 15/36 = 5/12 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes ..basic sample space analysis Creatorpk answered Mar 17 Creatorpk comment Share Follow See all 0 reply Please log in or register to add a comment.