Probability - tossing a coin until a head is obtained

A fair coin is tossed until a a head is obtained. So for example, the sample space can be modelled as {$H, TH, TTH, TTTH, ...$}. The probability measure is modelled such that $P(H)=\frac 12$, $P(TH)=\frac 14$, $P(TTH)=\frac 18$ etc.

Find the probability that the total number of tosses is:

(a) at least k

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What have you tried? –  Cameron Buie Oct 15 '13 at 10:25
So for (a), the probability of getting a head on the kth toss is 1/$2^k$ so I was trying to use this fact, although not sure how to calculate it as "at most k". –  Simon M Oct 15 '13 at 10:27
Have you been taught about the geometric distribution? That should be your first stop. Following on from there, you should try to make a bunch of infinite sums for each of the following parts, and evaluate those. The probability that the number of tosses, $t$ equals $1$ is known, the probability that $t=3$ is known, the probability that $t=5$ is known, and so on. What does this tell you about the probability that $t$ is odd? –  ymbirtt Oct 15 '13 at 10:34
Nope, I presume reading up about the geometric distribution will give me the knowledge for a solution? –  Simon M Oct 15 '13 at 10:37
It'll start you off on the first question, and give you a way of thinking about the others. –  ymbirtt Oct 15 '13 at 10:37

Let $X$ be the number of moves. Then $$P(X=k) = \frac{1}{2^k}$$ (a) $$P = 1 - P(X > k) = 1 - P(k \text{ tails}) = 1 - \frac{1}{2^k} = \frac{2^k - 1}{2^k}$$ (b) $$P = \sum_{k=0}^\infty P(X=2k+1) = \sum_{k=0}^\infty \frac{1}{2^{2k+1}} = \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k} = \frac{1}{2} \frac{1}{1- \frac{1}{4}} = \frac{2}{3}$$ (c) $$P = \sum_{k=0}^\infty P(X=3(k+1)) = \ldots = \frac{1}{7}$$ (d) $$P = P_b + P_c - \sum_{k=0}^\infty P(X = 3(2k+1)) = \ldots = \frac 2 3 + \frac 1 7 - \frac 1 8 \cdot \frac{1}{1 - \frac 1 {64}} = \frac{43}{63}$$
Thanks for this. Would c be $\frac 17$? –  Simon M Oct 15 '13 at 11:25
and then d be $\frac {43}{63}$? –  Simon M Oct 15 '13 at 11:31