Finding mean, variance and standard deviation from m.g.f. of coin flips

[gimpy]

I am having trouble with this question.

Let X equal the number of flips of a fair coin that are required to observe the same face on consecutive flips.
(a) Find the p.m.f. of X.
if found the p.m.f. to be $$f(x) = (\frac{1}{2})^{x-1}$$ for $$x=2,3,4,...$$

(b) Give the values of the mean, variance and standard deviation of X.
For this one i found the m.g.f. to be $$M(t) = E(e^{tx}) = \sum_{x \in S} e^{tx}f(x)$$
$$M^{'}(0) = xf(x) = E(X)$$ which is the mean.
Then
$$M^{''}(0) = x^{2}f(x) = E(X)$$
$$Var(X) = M^{''}(0) - [M^{'}(0)]^2$$

Is this correct?

Than after that i realized that 2 rolls of the dice was the minimum to get the same face on two consecutive flips. So i made $$S={1,2}$$ and evaluted them like this getting the Mean = 2 and Variance = 4 which is not correct.
What am i doing wrong? Do i have to use Infinite series or something?

I haven't even started on the standard deviation.

(c) Find the values of
$$(i) P(X \leq 3)$$
$$(ii) P(X \geq 5)$$
$$(iii) P(X = 3)$$

I can't start on these until i get part (b)

 

[matt grime]

S must be the set {2,3,4...}. yes you need to evalaute the infinite sums, but they aren't hard to do. the sum of nr^n and n(n-1)r^n are quite well known and you should be able to find them.

you can do c without doing b; it doens't involve moments at all.


probability it happens on the first throw is 0, probability on the second is 1/2, and on the third is 1/4 so the first probability is 3/4

x greater than 5 is 1- prob of on the 2nd 3rd or 4th, which is 1-1/2-1/4-1/8

on the fifth exactly
well one of HTHTT or THTHH must happen and the probability is...?