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].

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) = x P(TT)

              =   x (0.5 x 0.5) =

P(TT at the end of X tosses) =

similarly,

P(HH at the end of X tosses) =

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

P(X) = 2 x ( )

E(X) = Σx.2*( ) {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)()

E(y) = Σy.(y-1)() {y >= 2}

E(y) = 4

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: