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
