geometric distribution of mistrials between two hits

Question

I have the following Task to complete:

A Basketball Player has 10 tries. The table below shows the amount $x_k$ of mistrials till the $k^{th}$ success:

\begin{array}{|c|c|c|c|} \hline Trial \ k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline x_k & 1 & 0 & 2 & 3 & 0 & 1 & 0 & 0 & 2 & 2 \\ \hline \end{array}

Which distribution does the amount of mistrials $x_k$ have between two hits?

We got as far as to argue:

1. The event of X (the player has scored a hit) is bernoulli distributed; since $x \in \{1,0\} $
2. Since $ X_1, ...,X_n $ are independent repetitions of $X$, they should be geometrical distributed. (So $ X_1, ...,X_n $ is the Amount of failed attempts until the first hit)

Now I seem to be unable to find a way to continue the argument.

Are our assumptions correct?

Do I need to introduce another Event for the next hit to come?

Thanks in advance!

Answer 1

Yes, your assumptions are correct. The $x_k$-s are indeed geometrically distributed. Introducing another event would be wrong since you want to strictly study your given sample and its sample space.