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

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