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?

  • 1
    $\begingroup$ We aren't told anything about how the meals are distributed throughout the week. If we make the natural assumption, namely that each meal is equally likely to be served on any given day, then Veg$_1$ has a $\frac 15$ chance of being consumed on Wednesday as does Veg$_2$. Adding, we get $\frac 25$. $\endgroup$
    – lulu
    Apr 6, 2022 at 5:12

2 Answers 2


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.

  • $\begingroup$ Seriously? Even though there are more non-veg then veg, you think that it is far more likely that you'll get a veg meal on Wednesday? $\endgroup$
    – lulu
    Apr 6, 2022 at 5:30
  • $\begingroup$ You're right, I shouldn't be answering this late. I've corrected it. $\endgroup$
    – C. Ventin
    Apr 6, 2022 at 5:38
  • $\begingroup$ Post edit, this is correct but it's hard to generalize this method. If you prepared meals for the next $365$ days with exactly $17$ veg meals amongst the lot, the probability that the meal on day $193$ is a veg meal is $\frac {17}{365}$, though this would be hard to see if you tried to write out all the possible arrangements of the meals prior to that day. There are a lot of possible arrangements, after all. $\endgroup$
    – lulu
    Apr 6, 2022 at 5:41
  • $\begingroup$ Switching from P(vegetable | Wednesday) to P(Wednesday | vegetable) would make for a good additional answer. I wanted to reassure OP that the suggested method pans out too . . . $\endgroup$
    – C. Ventin
    Apr 6, 2022 at 5:44
  • $\begingroup$ @lulu I think that M.Reeves makes a good point here: "I wanted to reassure OP that the suggested method pans out too . . .". I also think that it is valuable to demonstrate to the OP, as M.Reeves answer does demonstrate, that his inelegant (i.e. cumbersome) but valid alternative approach does lead to the same final computation. $\endgroup$ Apr 6, 2022 at 23:06

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.

  • 2
    $\begingroup$ It's also apparent that if you stick the labels on first, but then you decide to eat Wednesday's meal on Monday and Monday's meal on Wednesday, there should be no change in the probability. $\endgroup$
    – user253751
    Apr 6, 2022 at 14:31
  • $\begingroup$ Although that does not generalize in and of itself, it relies on here proving mutual independence, which seems to go a bit beyond the scope of the question. $\endgroup$
    – C. Ventin
    Apr 6, 2022 at 14:45
  • $\begingroup$ @M.Reeves My answer does not assume mutual independence between the days of the week. Instead, my answer merely assumes that (in effect) on Sunday night, of the $5$ dinners prepared, $2$ of the $5$ dinners will be chosen at random to have the vegetables. That is, I am not assuming that the probability of vegetables on Wednesday is independent of whether there were vegetables on Monday. Instead, I am merely assuming that before any of the meals have been eaten, $2$ of the $5$ meals were randomly chosen to receive the vegetables. $\endgroup$ Apr 6, 2022 at 14:56
  • $\begingroup$ You're swapping P ( particular meal | particular day ) proportionally for P ( particular day | particular meal ). Just saying, that requires independence. It's not wrong, the problem seems to assume that. $\endgroup$
    – C. Ventin
    Apr 6, 2022 at 16:32
  • $\begingroup$ @M.Reeves I do not understand your comment. Please elaborate. What events are you explicitly referring to, and what independence between the events am I assuming. For example, if Event 1 was a coin flip coming up Heads, and Event 2 was a 6 sided die showing a $1$ face up, then you could say that I am assuming that Event 1 and Event 2 are independent events. Please clarify what assumptions that you think that I am making, re independent events, by very clearly identifying what you regard as Event-1, and Event-2, and what assumptions between these events that you think that I am making. $\endgroup$ Apr 6, 2022 at 16:42

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