1. $P\left(Y=1, X_{1}=1\right)$
- For $Y=1, \frac{X_{1}+X_{2}}{2}=1$ implies $X_{1}+X_{2}=2$, so $X_{1}=1$ and $X_{2}=1$.
- $P\left(Y=1, X_{1}=1\right)=P\left(X_{1}=1\right.$ and $\left.X_{2}=1\right)=p \cdot p=p^{2}$.
2. $P\left(Y=0, X_{1}=0\right)$
- For $Y=0, \frac{X_{1}+X_{2}}{2}=0$ implies $X_{1}+X_{2}=0$, so $X_{1}=0$ and $X_{2}=0$.
- $P\left(Y=0, X_{1}=0\right)=P\left(X_{1}=0\right.$ and $\left.X_{2}=0\right)=(1-p) \cdot(1-p)=(1-p)^{2}$.
Thus, the correct probabilities are:
$P\left(Y=1, X_{1}=1\right)=p^{2}$
$P\left(Y=0, X_{1}=0\right)=(1-p)^{2}$
The correct option is $\mathbf{B}$.
