I have recently come upon a probability question from this site (Binomial Distribution Problem - Desk problem) which goes as follows:
The mathematics department has 8 graduate assistants who are assigned to the same office. Each assistant is just as likely to study at home as in the office. How many desks must there be in the office so that each assistant has a desk at least 90% of the time?
For me the most confusing aspect of this question is the word "each" in the last sentence. Either the word "each" is to be interpreted as "all 8 assistants" (A) or as "any given one of the 8 assistants" (B).
Now in both case A and B we can break the problem up into scenarios in which $k$ assistants show up to the office and there are $m$ desks provided at the office. Since $0\leq{}k,m\leq{}8$ and $k$ is independent of $m$ it follows that there are 81 such scenarios. The probability $P(k)$ that $k$ assistants show up to the office is given by the binomial distribution $B(8,1/2)$:
$$P(k)=\binom{8}{k}\left(\frac{1}{2}\right)^k\left(\frac{1}{2}\right)^{8-k}$$
In both case A and B we also assume that every assistant who works at home has a desk. Now in case A the probability that all 8 assistants have a desk given that $k$ assistants show up to the office and there are $m$ desks provided at the office is:
$$P_A(k,m)= \begin{cases} 1&\text{if}\,{}k\leq{}m\\ 0&\text{otherwise}\\ \end{cases}$$
In case B the probability that any given one of the 8 assistants has a desk given that $k$ assistants show up to the office and there are $m$ desks provided at the office is:
$$P_B(k,m)= \begin{cases} 1&\text{if}\,{}k\leq{}m\\ [(8-k)+m]/8&\text{otherwise}\\ \end{cases}$$
It follows that in case A the probability $P_A(m)$ that all 8 assistants have a desk given that there are $m$ desks provided at the office is given by:
$$P_A(m)=\sum_{k=0}^{8}\left[P(k)*P_A(k,m)\right]$$
Similarly, the probability $P_B(m)$ that any given one of the 8 assistants has a desk given that there are $m$ desks provided at the office is:
$$P_B(m)=\sum_{k=0}^{8}\left[P(k)*P_B(k,m)\right]$$
Using these two formulas we obtain two different answers to the original question. In case A the minimum value of $m$ for which $P_A(m)\geq{}0.9$ is $m=6$ and in this scenario $P_A(6)\approx{}0.965$. In case B the minimum value of $m$ for which $P_B(m)\geq{}0.9$ is $m=4$ and in this scenario $P_B(4)\approx{}0.932$.
My question is twofold. Firstly, is my logic and math correct for both interpretations of the question? Secondly, which interpretation is the valid one?
Let $d$ be the number of desks, and let $N \sim \text{Binom}(8, 0.5)$ be the random variable representing the number of people who want desks at any given time. The question is asking us to find the minimum $d$ for which $$\Bbb{P}(N \leq d) \geq 0.9.$$ Since $$\Bbb{P}(N \leq d) = \sum_{i = 0}^d \Bbb{P}(N = i),$$ we can check this case-by-case. It so happens that $$\Bbb{P}(N \leq 6) = 96.48 \text{%, but } \Bbb{P}(N \leq 5) = 85.55 \text{%},$$ so the magic number is $d = 6$ desks to give each student who wants one a desk at least 90% of the time.
Edit: From the definition of "each" as more-or-less synonymous with "every", we can see that any time even one assistant doesn't get a desk, we are not in a case where "each assistant got a desk", and so it seems more appropriate to read the question as Case A, where all assistants must get a desk, than Case B, considering an individual assistant.
If they meant Case B, it probably would have been clarified by wording it to refer to a specific and maybe even named assistant (e.g. "Henry is one of 8 assistants in an office with $d$ desks, each of whom chooses to work at home or the office with 50-50 probability. Every assistant's choice is independent of the others. What is the probability that Henry has a desk at least 90% of the time?")
Ultimately the wording is ambiguous and either choice can be defended, but it seems like the event whose probability is 90% should really be "each assistant gets a desk" based on the exact wording of the question, rather than "any particular assistant gets a desk".