2 votes 2 votes HeadShot asked Sep 3, 2018 HeadShot 346 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply MiNiPanda commented Sep 3, 2018 i reshown by MiNiPanda Sep 3, 2018 reply Follow Share I am getting option A.. 0 votes 0 votes MiNiPanda commented Sep 3, 2018 reply Follow Share If the first toss is H then all the remaining (n-1) tosses have to be H. If 2nd toss is H then remaining (n-2) tosses have to be H. All tosses cannot be T as given that atleast one toss has to be H so out of n tosses the last toss must be H. P(HH..Hn times H ) + P(THH...H(n-1) times H) + P(TTHH...H(n-2) times H) + ........+P(TTTT..TH1 time H) How many such terms are there? n [1<=count(H)<=n] =(2-n) + (2-n) ....... (2-n)n times = n*2-n 3 votes 3 votes arvin commented Sep 3, 2018 reply Follow Share let n=1 sample space = (H) ,(T) favourable event =(h) ---------> probability = 1/2 n=2 sample space= (HH) (HT) (TH) (TT) favourable outcome = (HH)(TH) --------> probability = 2/4 = 1/2 n=3 sample space = 8 events fav- outcome= (HHH), (THH),(TTH) =3 Probability = 3/8.... all these matches with option A... answer 2 votes 2 votes MiNiPanda commented Sep 3, 2018 reply Follow Share @arvin after a long time! 0_0 :P 0 votes 0 votes arvin commented Sep 3, 2018 reply Follow Share @minipanda : was stucked in personal life havocs :l lol 0 votes 0 votes MiNiPanda commented Sep 3, 2018 reply Follow Share Haha..my time is also coming soon.. :( 0 votes 0 votes srestha commented Sep 3, 2018 reply Follow Share option are of new type, in so simple question :) 0 votes 0 votes Please log in or register to add a comment.