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

Answers

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

number of timesthat the sequenceHH appears=9/4Xfollows abionomial distributionwith success probabilityp=1/4andn=9the number of trials.let X = HHappeared in a flip coin of10 tosses.HH twicein a row has probability equal top=1/4.HHis yoursuccessoutcome and the rest {HK,KH,KK} consists of the failure outcomes.HHin10 tosses=9 waysX∼Bionomial (9,1/4).E[X]=np=9/4coin tossesfrom the given link