Finding $P(3≤X≤4\mid Y=2)$

Question

Let $X$ be a number chosen according to the discrete uniform distribution on the set $S$ of $10$ decimal digits ($0$ through $9$) and let $Y$ be a number chosen according to the discreet uniform distribution on $S$ with $X$ removed.

1. Determine and identify the conditional PMF of $Y$ given $X=x$.
2. Determine and identify the conditional PMF of $X$ given $Y=y$.
3. Find $P(3\le X\le4\mid Y=2)$

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1. From what I know I have $P(X=x\mid Y=y)=1/10$.
2. And $P(Y=y \mid X=x)=\dfrac19$.
3. $$P(3\le X\le4\mid Y=2)={P(3\le X\le4 \cap Y=2)\over P(Y=2)}={\displaystyle \sum_{i=3}^4\frac1{10}\cdot\frac19 \over \dfrac19}=\frac15.$$

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I am practically sure of everything except part 3. If I could get some tips It would be greatly appreciated.

Answer 1

You have answers 1 and 2 mixed up, and one not quite correct.   Answer 3 is way off.

1. The conditional PMF for $Y$ given $X=x$ is $\bbox[0.5ex]{\mathsf P(Y{=}y\mid X{=}x) = 1/9}$, though you should add that the support is $y\in S\setminus\{x\}$ for all $x\in S$.

2. The conditional PMF for $X$ given $Y=y$ is $\bbox[0.5ex]{\mathsf P(X{=}x\mid Y{=}y)}$ but this is not equal to $1/10$.   As a hint, notice that the support will be $x\in S\setminus\{y\}$ for all $y\in S$, because when given that you will select $Y=y$ you cannot select $X=y$, but you may select from the other nine values without bias.

3. $\bbox[0.5ex]{\mathsf P(3{\leq}X{\leq}4\mid Y{=}2)=\mathsf P(X{=}3\mid Y{=}2)+\mathsf P(X{=}4\mid Y{=}2)}$.   There is no need to use Bayes' Rule if you already have the conditional PMF.