Suppose that $x_1, x_2,...$ are independent copies of random variable $\xi$ having distribution $P(x=1)=P(x=-1)=\frac{1}{2}.$ Let $S_0=0$, $S_k=x_1+x_2+...+x_k$, $k \geq1,$ and $a,b$ be non negative integer numbers such that $a-b>1$. Find probabilities $$P(S_2=2, S_{a+b}=a-b), \text{ } P(S_2=-2, S_{a+b}=a-b).$$
I know that $P(S_2=2)=P(x_1=1, x_2=1)=\frac{1}{2}*\frac{1}{2}$ and $P(S_2=-2)=P(x_1=-1, x_2=-1)=\frac{1}{2}*\frac{1}{2}.$ But what about integers $a$ and $b$. What should I do with them?
You can start with: $$P(S_2=2,S_{a+b}=a-b)=P(S=2)P(S_{a+b}=a-b\mid S_2=2)=\frac14P(S_{a+b-2}=a-b-2)$$
Now the question rises: how many ways are there from $0$ to $a-b-2$ taking $a+b-2$ steps?
If there are $r$ steps to right and $l$ to left then we have the equalities:
leading to $r=a-2$ and $l=b$ and showing that there are $\binom{a+b-2}{b}$ routes.
I leave the rest to you.
Edit:
If e.g. $a=3$ and $b=2$ then:
$\begin{aligned}P\left(S_{a+b}=a-b\mid S_{2}=2\right) & =P\left(S_{5}=1\mid S_{2}=2\right)\\ & =P\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1\mid x_{1}+x_{2}=2\right)\\ & =P\left(2+x_{3}+x_{4}+x_{5}=1\mid x_{1}+x_{2}=2\right)\\ & =P\left(x_{3}+x_{4}+x_{5}=-1\mid x_{1}+x_{2}=2\right)\\ & =P\left(x_{3}+x_{4}+x_{5}=-1\right)\\ & =P\left(x_{1}+x_{2}+x_{3}=-1\right)\\ & =P\left(S_{3}=-1\right)\\ & =P\left(S_{a+b-2}=a-b-2\right) \end{aligned} $
The $5$-th equality rests on independence.
The $6$-th equality on the fact that $x_1+x_2+x_3$ and $x_3+x_4+x_5$ have the same distribution.