A fair coin is tossed till two heads occur in succession. What is the probability that the number of tosses required is less than 6?

Explanation

We need the probability of needing 2, 3, 4, or 5 tosses. The successful sequences are HH (for n=2, prob 1/4), THH (for n=3, prob 1/8), TTHH and HTHH (for n=4, prob 2/16), and TTTHH, HTTHH, THTHH (for n=5, prob 3/32). Summing these probabilities yields \frac{1}{4} + \frac{1}{8} + \frac{2}{16} + \frac{3}{32} = \frac{8+4+4+3}{32} = \frac{19}{32}.

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