Question 1160576
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The problem asks to find the probability to have a head <U>UNDER THE CONDITION</U> that <U>the three first tosses give a tail</U>.


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) = {{{1/2^4}}};  

             P(5) = {{{(1/2)}}}.P(TTTTH) = {{{(1/2)*(1/2^5)}}} = {{{1/2^6}}};

             P(6) = {{{(1/2^2)}}}.P(TTTTTH) = {{{(1/2^2)*(1/2^6)}}} =  {{{1/2^8)}}},   and so on . . . 


So,  P is the sum of the infinite geometric progression with the first term  {{{1/2^4}}}  and the common ratio of {{{(1/2)*(1/2)}}} = {{{1/4}}}.

The sum of this progression is  P = {{{0.5^4/(1-1/4)}}} = {{{0.5^4/((3/4))}}} = {{{0.5^4*4/3}}} = {{{0.25/3}}} = 0.083333...   


It is the <U>ANSWER</U> to this problem.
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Solved.


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Notice that my answer is different from the solution by &nbsp;Edwin, &nbsp;and the logic is &nbsp;DIFFERENT, &nbsp;too.