What is the equivalent probability of $P(X + Y = 11 | X=6)$?

Question

Let $X \sim Bin(10, .5)$, $Y \sim Bin(10, .5)$ and $X$, $Y$ are independent.

Which probability $P(X=6) \cdot P(Y=5)$, or $P(Y=5)$ (since X is given) is an equivalent probability of $P(X + Y = 10 | X=6)$ and why?

Answer 1

$$\begin{align}\mathsf P(X+Y=10\mid X=6)&=\dfrac{\mathsf P(X+Y=10, X=6)}{\mathsf P(X=6)}&&\small\text{definition of conditional probability}\\[2ex]&=\dfrac{\mathsf P(Y=4, X=6)}{\mathsf P(X=6)}&&\small\text{obviously}\\[2ex]&=\dfrac{\mathsf P(Y=4)\,\mathsf P(X=6)}{\mathsf P(X=6)}&&\small\text{independence of }X, Y\\[2ex]&=\mathsf P(Y=4)&&\small\text{cancelation}\end{align}$$

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PS: your body text says $X+Y=10$ where as the title said $X+Y=11$. So...