=1/2
P[TH]=1/4
P[TTH]=1/8
P[TTTT]=P[TTTH]=1/16
let B the event that head did not appear on either of the first 2 tosses,then B is known as reduced sample space and it is:
B={TTH,TTTT,TTTH}
let A be the event that the coin was tossed 4 times.then A contains the sample points {TTTT,TTTH}
therefore P[A|B]=P[AB]/P[ B]
now AB={TTTH,TTTT},P[AB]=1/16 + 1/16= 1/8
and P[ B]=1/8 + 1/16 + 1/16 =1/4
P[A|B]=(1/8) / (1/4) =1/2
(1) thus the problem and solution ends.but the problem is on the 2nd line of the solution (enclosed by /* */),solver took {H,TH,TTH,TTTH,TTTT} as sample space.why we should take TTTT as sample space when we are given that we should count until the head appears?
(2)theres a next part of this problem,where it was told to find the probability of appearing head
in just 3 tosses with previous conditions.there the solver took another sample point C={TTH},why he did not took {TTH,TTT} instead like the previous one...?
i am pretty dumb in using latex,pls forgive me for that....