Let $(X_k)_{k = 1, 2, 3}$ be a sequence of i.i.d. random variables. Let $Y_k = X_k + X_{k - 1}$ with $\mathbb{P}(X_k = 1) = \mathbb{P}(X_k = 0) = \frac{1}{2}$.
Show that $$\mathbb{P}(Y_3 = 0 \mid Y_2 = 1) = \frac{1}{4}.$$
My attempt: We have the conditional probability $$\mathbb{P}(Y_3 = 0 \mid Y_2 = 1) = \frac{\mathbb{P}(Y_3 = 0, Y_2 = 1)}{\mathbb{P}(Y_2 = 1)},$$ where we can substitute $Y_3 = X_3 + X_2$ and $Y_2 = X_2 + X_1$. So, $$\mathbb{P}(Y_3 = 0 \mid Y_2 = 1) = \frac{\mathbb{P}(X_3 + X_2 = 0, X_2 + X_1 = 1)}{\mathbb{P}(X_2 + X_1 = 1)}.$$ But now, I'm stuck. How do we compute the probability of a sum of two random variables? Do we say: $$\mathbb{P}(X_2 + X_1 = 1) = \mathbb{P}(X_2 = 0, X_1 = 1) + \mathbb{P}(X_2 = 1, X_1 = 0),$$ and since we have independence get $$\mathbb{P}(X_2 = 1, X_1 = 0) = \mathbb{P}(X_2 = 1) \cdot \mathbb{P}(X_1 = 0) = \frac{1}{4}?$$
Yes, your computation pf the denominator is correct. And the numerator is even simpler: $P(X_3+X_2=0, X_2+X_1=1)=P(X_3=0,X_2=0,X_1=1)=\frac 1 8$.