If we toss a fair coin until we get two consecutive heads or two consecutive tails, let X be the number of tosses. If you toss a fair coin

Question

If we toss a fair coin until we get two consecutive heads or two consecutive tails, let X be the number of tosses. If you toss a fair coin until you get a tail that is preceded by a head, let Y be the number of tosses. (1). Find the PMF of X and E(X). (2). Find the PMF of Y and E[Y].

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Valentina 7 months 2021-10-07T22:44:47+00:00 1 Answer 0 views 0

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    2021-10-07T22:46:26+00:00

    Answer:

    1)  PMF of X and E(X) is 3

    2) PMF of Y and E[Y] is 4

    Step-by-step explanation:

    (1). Find the PMF of X and E(X)

    it has to have TT and HH at end

    if at end there is TT then the sequence has to be : …….THTHTHTT

    if at end there is HH then the sequence has to be : …….HTHTHTHH

    no. of tosses before last two = x-2

    these x-2 tosses also have fix pattern so,

    P(……TT) = [tex]0.5^{x-2}[/tex] x P(TT)

                  =  [tex]0.5^{x-2}[/tex] x (0.5 x 0.5) = [tex]0.5^{x}[/tex]

    P(TT at the end of X tosses) = [tex]0.5^{x}[/tex]

    similarly,

    P(HH at the end of X tosses) = [tex]0.5^{x}[/tex]

    P(X) = P(TT at the end of X tosses) +  P(HH at the end of X tosses)

    P(X) = 2 x ([tex]0.5^{x}[/tex]
    )

    E(X) = Σx.2*([tex]0.5^{x}[/tex]
    ) {x >= 2}

    E(X) = 3  

    2.  Find the PMF of Y and E[Y].

    For the y-2 tosses before the HT if there is any heads then it will only be succeeded by heads till we reach the HT

    therefore the y-2 tosses before HT are of the form : TkH(y-2)-k

    k can be 0 to y-2 therfore y-2 +1 possibilities

    k can have y-1 values

    P(y) = (y-1)([tex]0.5^{y}[/tex])

    E(y) = Σy.(y-1)([tex]0.5^{y}[/tex]) {y >= 2}

    E(y) = 4

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