Binomial/Geometric/Bayes perspectives on coin tosses?

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So I have the following question which I am trying to figure out/verify answers.

a) I used the binomial probability mass function with n= 10 and p = 0.5 to determine the values. I think a success can be thought of as getting a head, as denoted by k. So I got,

$$P( k = 5 ) = 0.24609375 \qquad P( k = 10) = 0.0009765625$$

Is this approach correct ?

b) For this part, I calculated $P( X > 1 )$ by doing the following:

\begin{align} P(X > 1 ) &= 1 - P ( X = 1 ) \\ &= 1 - 0.5 ( 0.5 ) ^ { 1 - 1 } = 0.5 \end{align}

For $P( X > 10 )$, how would I calculate that. Would it be simply $1 - P(X = 10)$ or do I have to add up all the probabilities up $X = 10$, as in:

$$P( X > 10 ) = 1 - [ P( X = 1) + P ( X = 2) + \cdots + P( X = 10 ) ?$$

c) I have very little idea on what to do here since the $p > 1/2$ which is not an exact number.

I am also trying to understand how calculating these values help me settle the dilemma. Any help would be appreciated.