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The problem asks to find the probability to have a head UNDER THE CONDITION that the three first tosses give a tail.
It is the sum of probabilities
P = P(4) + P(5) + P(6) + . . .
where P(4) is the probability to get H first time at the 4-th toss;
P(5) is the probability to get H first time at the 5-th toss, assuming that the outcome was T at the 4-th toss;
P(6) is the probability to get H first time at the 6-th toss, assuming that the outcome was T at the 4-th and 5-th toss,
and so on . . . (infinite sum)
Notice that P(4) = P(TTTH) = ;
P(5) = .P(TTTTH) = = ;
P(6) = .P(TTTTTH) = = , and so on . . .
So, P is the sum of the infinite geometric progression with the first term and the common ratio of = .
The sum of this progression is P = = = = = 0.083333...
It is the ANSWER to this problem.
Solved.
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Notice that my answer is different from the solution by Edwin, and the logic is DIFFERENT, too.