Suppose you toss a fair coin 10 times resulting in a sequence of heads (H) and tails (T). Let X be the number of times that the sequence HH appears. For example, HH appears thrice in T HT HHT HHHT


  1. The number of times that the sequence HH appears = 9/4
    X follows a bionomial distribution with success probability
    p=1/4 and n=9 the number of trials.
    Too see this let X = HH appeared in a flip coin of 10 tosses.
    Clearly, as you said to get HH twice in a row has probability equal to p=1/4.
    So, you look at your problem from the point of view that HH is your success outcome and the rest {HK,KH,KK} consists of the failure outcomes.
    Now possible ways you have in order to get HH in 10 tosses = 9 ways
    in order to figure this out you look at the positions where the first H can appear in a sequence of HH.
    It can appear in any of the first 9 positions.
    So, we end up that X∼Bionomial (9,1/4).
    And this has E[X]=np=9/4
    To learn more about coin tosses from the given link


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