computing probabilities of an negative binomial random variable Would appreciate if you could check if I answered the questions correctly:
$X$ is a Negative Binomial random variable with the parameters $\frac{1}{2}$ and $r =1,2,3,\ldots$.
$Y$ is a random variable that is defined as: $Y = 
\begin{cases}
X,  & x \le r+1 \\
X-1, & x \ge r+2
\end{cases}$

(1) Find $\Pr\{ Y = r+1 \}$ 

The result I got: $r\,(\frac{1}{2})^{r+2}$. Since we need, according to the definition of Negative Binomial, to calculate the probability of trials to make $r$ success, then we can calculate it by multiplying the parameters ($1/2$ and $r$) times $r+1+1$ to cover it all.

(2) Find $\mathbb{E}[Y]$

The result I obtained is $2r-1-\frac{r+2}{2^r}$. I calculated it using $r/p$ while trying to cover all of $Y$.
Hope I did this one correctly. Would appreciate your comments or corrections.

Your description in (1) and the fact that you quoted $\mathbb{E}[X] = r/p$ for (2), I take it that you adopt the total number of trials for the definition of $X$ the Negative Binomial.
\begin{align*}
\Pr\bigl\{ Y = r+1 \bigr\} &=  \overbrace{\Pr\bigl\{X = r+1 \bigr\} }^{Y=X} +  \overbrace{ \Pr\bigl\{ X = r+2\bigr\} }^{Y=X-1} \qquad\text{, by definition of $Y$} \\
&= {r\choose r-1} p^rq + {r+1\choose r-1} p^r q^2 \Bigg|_{p=q=1/2}\\
&= r \cdot \frac1{ 2^{r+1}} + \frac{(r+1)r}{2!} \cdot \frac1{ 2^{r+2}} \\
&= \frac{ 4r }{ 2^{r+3}} + \frac{ r^2 + r }{ 2^{r+3}}\\
&= \frac{ r(r+5) }{ 2^{r+3}}
\end{align*}
Similarly, the expectation splits into two parts.
\begin{align*}
\mathbb{E}[ Y ] &= \Bigl\{ \text{contribution from $X \leq r+1$} \Bigr\} + \Bigl\{ \text{from $X \geq r+2$} \Bigr\} \\
&= \sum_{k=r}^{r+1} \Bigl[ \color{magenta}k \cdot \Pr\bigl\{ X = k \bigr\} \Bigr] + \Biggl\{ \mathbb{E}[ X-1 ] - \sum_{k=r}^{r+1} \Bigg[ (\color{magenta}k-1) \cdot \Pr\bigl\{ X = k \bigr\} \Bigg]  \Biggr\} \\
\end{align*}
After canceling the $\color{magenta}{\text{magenta }k}\,$, we have
\begin{align*}
\mathbb{E}[ Y ] &= \sum_{k=r}^{r+1} \Bigl[ 1 \cdot \Pr\bigl\{ X = k \bigr\} \Bigr] + \mathbb{E}[ X-1 ]\\
&= \frac1{ 2^r} + \frac{ r }{ 2^{r+1}} + \frac{r}p\Bigg|_{p = 1/2} - 1 \\
&= 2r-1 + \frac{ r+2 }{ 2^{r+1}}
\end{align*}
Your own answer is just off by a minus sign.
