Tossing a Fair Coin: Probability Mass Function

[VantagePoint72]

Homework Statement

A fair coin is tossed repeatedly and X is the number of tosses before the first head appears. You independently repeat the experiment, and Y is the number of tosses before the first head appears in the second sequence of tosses.
a. Give the probability mass function of X
b. Find P(X>n), for n≥1
c. Find P(X=Y)
d. Find P(X>Y)

Homework Equations

Geometric pmf: p(x) = p(1-p)^x-1

The Attempt at a Solution

a. I believe this should be P(X=x) = (1/2)^x+1. Is this correct so far?

b. P(X>n) = 1 - P(X≤n) = 1 - \sum \limits_{k=1}^n(1/2)^x+1

I feel like a closed form for this is expected. I know there's some kind of formula for sums of powers, but we never looked at it in class. So, I'm wondering if there's another way to do this question.

c. P(X=Y) = \sum \limits_{k=1}^∞(1/2)^2x+2

Is this correct? If so, same issue with explicitly evaluating the sum as with b.

d. Since the problem is symmetric with respect to X and Y, presumably the answer is:
P(X>Y) = 0.5*[1-P(X=Y)]
Thus, I need the closed form solution from c. This solution relies on the fact that X and Y have the same distribution. Is there a more general method for this?

 

[Dick]

The closed form solution for the sum of powers is just the geometric series. See http://en.wikipedia.org/wiki/Geometric_series

 

[VantagePoint72]

Baha oh my head is stupid tonight, it seems. Aside from that, are my answers correct?

 

[Ray Vickson]

Baha oh my head is stupid tonight, it seems. Aside from that, are my answers correct?

Yes, except for your not defining what, exactly, you mean by x and what is its range. You see, x is different in the geometric distribution p(x) = p(1-p)^x-1 and in your P(X=x) = (1/2)^x+1. Defining what your symbols mean is an important part of the solution process.

RGV

 

[VantagePoint72]

Good point, I've let myself get a little sloppy. Thanks Dick and Ray!