Calculating the probability of a specific event knowing only the expected value, without distribution

Question

I'm having some trouble with the following exercise question.

• An office parking lot provides 12 parking spots.
• There are 14 employees.
• On average, 10 employees arrive by car.
• What is the probability that parking space is sufficient on a given working day?

My attempt so far:

As I understand, "on average" is equivalent to the expected value. Let $X$ denote the number of employees arriving on a day. Then,

$$E (X) = \sum_{i=0}^{14} p_i x_i = 10$$

So the distribution isn't uniform, since then the expected value would be 7. Also, this implies $P(X \leq 10) = P(X \geq 10)$, although I don't see how that's useful.

What I need to calculate is $$P(X \leq 12) = \sum_{i=0}^{12} p_i x_i = 1- \sum_{i=13}^{14} p_i x_i$$

Now, without knowing the distribution, I'd say there is no way to obtain the correct solution.

• For instance, it could be that $p_{10} = 1$ and then $P(X \leq 12) = 1$.
• It could also be that $p_{6} = p_{7} = p_{13} = p_{14} = 0.25$ and then $P(X \leq 12) = 0.5$

I also don't see any point in assuming a normal distribution in this case.

• If it were centered on the median value, the expectation would be 7 cars instead of 10 per day.
• If it were centered on 10, the expectation would be less then 10, because $p_i = 0$ for $i > 14$.
• It could work if it were centered somewhere between 10 and 14, but what sense does that make?

Am I missing something?

Answer 1

I think is that the question implies some assumptions. For instance, the fact that you arrive at an expected value of $7$ is only because you assumed the probability of an individual taking a car to be identical to that of a fair coin; instead, just take $p=\frac{5}{7}$, that leads you to an answer and I would guess that this is the one they are looking for.

However, as a side note, I wanted to mention that it's useful to think of concentration inequalities. These won't give you exact answers but can get you fairly tight bounds depending on the information you have at hand. For instance, you can use Markov's inequality here; that's not a great bound for this, so it's not particularly useful.

Answer 2

I think this is a disguised binomial distribution - that the unstated assumption is that on each day each of the $14$ employees drives to work independently with the same probability.

The fact that $10$ do so on average allows you to compute the probability.

Then you can figure out the probability that more than $12$ will drive.