Two questions about Independence of two RVs (building understanding) Situation A:
You are a math teacher writing a word problem for a quiz. You start off by stating,


*

*$X \sim Uniform(0,1)$

*$Y \sim Uniform(1,2)$

*$X$ and $Y$ are dependent
My first question: Does the last bullet point above make any sense? I don't think so... if I was writing a quiz question I couldn't use all the above bullet points at the same time. My thinking is the distributions of $X$ and $Y$ are fully known and independent of eachother. Therefore $X$ and $Y$ can never be dependent.

Situation B:


*

*$X \sim Uniform(0,1)$

*$Y = X+1$

*Clearly $X$ and $Y$ are dependent since $X$ and $X+1$ are dependent.
My second question: What If I say, since $Y=X+1$ which means I fully know the unconditional distribution of $Y$ and that is $Y \sim Uniform(1,2)$, and therefore $X$ and $Y$ are independent. I know this isn't fair to say, because give you know $X$ that means you know $Y$, but what is wrong with the "fully know the distribution" argument above?
Thanks for your help and patience.
 A: Comment: There are infinitely many joint distributions on
the rectangle with vertices at $(0,1)$ and $(1,2)$
that have your specified uniform marginal distributions, and yet
are not independent. 
An important principle here is that it is impossible, in general, to find the joint distribution corresponding to two marginal distributions (without additional information). However, if you know the two
marginal distributions are independent, then
the joint distribution can be found by multiplication.
In the upper half of the plot below (based on a simulation for
convenience in plotting), I illustrate an example where independence fails. Your example with
$X \sim \mathsf{Unif}(0,1)$ and $Y = X+1$ has what is sometimes called a
'degenerate joint distribution' with all probability
on a line. It is illustrated at the bottom of the figure.
In both cases, $X$ and $Y$ cannot be independent because
$P(X < 1/2) = P(Y < 3/2) = 1/2,$ but
$P(X < 1/2,\,Y < 3/2) = 1/2 \ne 1/4.$

Note: The R code for making the figure is shown below
in case anyone is interested.
par(mfrow=c(2,3));  m = 50000

 x = runif(m, 0, 1);  v = runif(m, 1 ,1.5)
 y = v;  y[x>1/2] = v[x>1/2]+ .5
   plot(x,y, pch=".")
   hist(x, br=10, prob=T, col="skyblue2")
   hist(y, br=10, prob=T, col="skyblue2")

 x = runif(m, 0, 1);  y = x+1
   plot(x,y, pch=".")
   hist(x, br=10, prob=T, col="skyblue2")
   hist(y, br=10, prob=T, col="skyblue2")

par(mfrow=c(1,1))

