<s>[INST] The probability that it will take at least $k$ tosses until the first head A fair coin is tossed until a head is obtained. So for example, the sample space can be modeled as {$H, TH, TTH, TTTH, ...$}. The probability measure is modeled so that $P(H)=\frac 12$, $P(TH)=\frac 14$, $P(TTH)=\frac 18$ etc.
Find the probability that the total number of tosses is at least $k$.
Progress: the probability of getting first head on the $k$th toss is $1/2^k$, so I was trying to use this fact, although not sure how to calculate it as "at least k".
 [/INST] 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) $$P = P_b + P_c - \sum_{k=0}^\infty P(X = 3(2k+1)) = \ldots = \frac 2 3 + \frac 1 7 - \frac 1 8 \cdot \frac{1}{1 - \frac 1 {64}} = \frac{43}{63}$$
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