A coin is tossed twice, if Ω={HH,HT,TH,TT}
and Borel Sigma algebra = 2^3 = { {1}, {1, 2}, {1, 2, 0}, {2}, {1, 0}, ∅, {0}, {2, 0} } then {1, 2} corresponds to {HH, HT, TH} and {1, 2) then how/why on real line it has the interval (0.75, 6)

Question

A coin is tossed twice, if Ω={HH,HT,TH,TT}
and Borel Sigma algebra = 2^3 = { {1}, {1, 2}, {1, 2, 0}, {2}, {1, 0}, ∅, {0}, {2, 0} } then {1, 2} corresponds to {HH, HT, TH} and {1, 2) then how/why on real line it has the interval (0.75, 6)

anonymous · Answer

is it because HH,HT, & TH are total 6 outcomes in {HH,HT,TH} (counting as HH=HT=TH=2) so that's why 6 is in (0.75, 6), and in Ω there are 8 outcomes (again counting as HH=HT=TH=TT=2, and total 8) & (chosen set of selected outcomes)/(total outcomes) i.e. 6/8 = 0.75 and now that's why 0.75 is in (0.75, 6). Similarly we can calculate it like as 3 outcomes have been chosen from Ω i.e. HH,HT,TH so 3/4 = 0.75 & that's why 0.75 is in (0.75, 6) and as each outcome consists of 2 possible values therefore 3*2=6 & that's why 6 is in (0.75, 6) ??????

anonymous · Answer

@amistre64 @mathslover @Zarkon @TuringTest @KingGeorge @satellite73  please help me