SOLUTION: A fair coin is tossed repeatedly until a head is obtained. The probability that the coin has to be tossed at least four times is...........................................? [Give

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Question 1160576: A fair coin is tossed repeatedly until a head is obtained. The probability that the coin has to be tossed at least four times is...........................................?
[Give your answer to nearest 3 decimal places.]

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
1-P(H)-P(TH)-P(TTH) = 1-0.5-0.5²-0.5³ = 1-0.5-0.25-0.125 = 0.125

Edwin

Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
.

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) = 1%2F2%5E4;  

             P(5) = %281%2F2%29.P(TTTTH) = %281%2F2%29%2A%281%2F2%5E5%29 = 1%2F2%5E6;

             P(6) = %281%2F2%5E2%29.P(TTTTTH) = %281%2F2%5E2%29%2A%281%2F2%5E6%29 =  1%2F2%5E8%29,   and so on . . . 


So,  P is the sum of the infinite geometric progression with the first term  1%2F2%5E4  and the common ratio of %281%2F2%29%2A%281%2F2%29 = 1%2F4.

The sum of this progression is  P = 0.5%5E4%2F%281-1%2F4%29 = 0.5%5E4%2F%28%283%2F4%29%29 = 0.5%5E4%2A4%2F3 = 0.25%2F3 = 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.