What is the Probability of Eating a Certain Meal on a Given Day? This is a problem that was given during a discussion section in the first week of my statistics class that I might be overthinking and misunderstanding.

You prepare 5 meals for the week, 2 with vegetables and 3 without. Starting on Monday, a meal is consumed each day until Friday. What is the probability that you will eat a meal with vegetables on Wednesday?

The answer to this problem was simply: $$\frac{(\text{# of Vegetable Meals})}{(\text{Total # of Meals})}$$ or $\frac{2}{5}$.
My confusion stems from, if a meal is consumed each day wouldn't the number of meals that we can choose from get smaller as we near the end of the week? So by Wednesday there would be only 3 meals to pick from. In addition, why do we not have consider the 3 different cases of the meals eaten before Wednesday? If a vegetable meal is eaten on Monday and Tuesday, then there would be a 0 chance of eating one on Wednesday. What about the cases where there was already 1 vegetable meal eaten before Wednesday? Would that not make the probability of eating a vegetable meal on Wednesday be: $$P(\text{Vegetable Meal on Wednesday)} = \frac{2}{5}*\frac{3}{4}*\frac{1}{3}$$ or if no vegetable meals are eaten before Wednesday at all: $$P(\text{Vegetable Meal on Wednesday})=\frac{3}{5}*\frac{2}{4}*\frac{2}{3}$$
Why would we not sum up these prbabilites to get the actual probability of eating a vegetable meal on Wednesday?
 A: The probability changes depending on what was eaten in the prior days. Say $v$ is a meal with a vegetable, $n$ is one without.
The possible combinations for Monday and Tuesday are:

*

*$vv$: $P = \frac 2 5\cdot\frac 1 4$, = $\frac 1 {10}$

*$vn$: $P = \frac 2 5\cdot \frac 3 4$, = $\frac 3{10}$

*$nv$: $P = \frac 3 5\cdot \frac 2 4$, = $\frac 3{10}$

*$nn$: $P = \frac 3 5\cdot \frac 2 4$, = $\frac 3{10}$
In each case, what is the chance of $v$ on Wednesday?

*

*$vv\mapsto v$: $P = 0$

*$vn\mapsto v$: $P = \frac 1 3$

*$nv\mapsto v$: $P = \frac 1 3$

*$nn\mapsto v$: $P = \frac 2 3$
To account for each case, weight the probability of $v$ on Wednesday according to the likelihood of the given Monday-Tuesday scenario, then combine:
$P = \frac 1{10}\cdot 0 + \frac 3{10}\cdot\frac 1 3+\frac 3{10}\cdot\frac 1 3 + \frac 3{10}\cdot\frac 2 3$, $= \boxed{\frac {2}{5}}$
Why does this work?
Well, the probability of any event (such as eating vegetables Wednesday) equals the probability of the event occurring in any possible scenario, combined. For example: You go to work on rainy days and on sunny days. So the number of days you go to work is the number of days you go to work when it rains, plus the number of days you go to work when it is sunny.
So for eating vegetables Wednesday, the probability is the same as the probability of eating vegetables Wednesday combined for each possible Monday-Tuesday scenario.
$\mathrm P(v) = \mathrm P(vv\mapsto v) + P(vn\mapsto v) + P(nv\mapsto v) + P(nn\mapsto v)$
I know it can be confusing learning how to work with probabilities. In my experience, introductory class questions were designed to trick the student rather than focusing on developing strong intuition or modeling ability. Please do feel free to follow up in the comments if having further questions or concerns.
A: Alternative perspective:
You can assume, without loss of generality that the meals are all prepared, in advance, on Sunday night.  Further, you can similarly assume that on Sunday night, you then stick a label on each of the $5$ meals, with the label identifying which meal will be eaten on which day.
Note that the $2$ assumptions in the above paragraph do not have any effect on the probability of eating vegetables on Wednesday.  That is, the probability is unaffected by whether Wednesday's meal is prepared and assigned on Sunday night, rather than Tuesday night or Wednesday morning.
Then, it is easy to see that the probability of having vegetables on Wednesday must be the same as the probability of having vegetables on Monday.
