Alaska license plates have two letters followed by three numbers. What is the probability that a randomly chosen license plate will have an NC with the number ending in a 3?
License plate probability question
probability
Related Solutions
No. At the very least, you also need to know the a priori probability that a plate is a vanity plate. That is, when you ask:
is there a way to determine the probability that a license plate with no digits was randomly chosen and is not a vanity plate?
you are asking for the conditional probability $P(V|N)$ with $V$ being the event of the plate being a vanity plate, and $N$ being the event of the plate having no digits.
Now, you could of course try to use Bayes' formula:
$$P(V|N)=\frac{P(N|V)\cdot P(V)}{P(N)}=\frac{1\cdot P(V)}{P(N)}= \frac{ P(V)}{P(N)}$$
But, like I said, you'd need to know the a priori probability $P(V)$ to use this.
Moreover, the $P(N)$ that this formula refers to is not the same as the $P(N)$ that you calculated: the $P(N)$ you calculated was the probability of getting no digits on a plate where all symbols are picked randomly, i.e. this was assuming no one has requested a vanity plate!
Now, here is something you can do: find the actual $P(N)$, i.e. find the percentage of plates that are actually out there being used that have no numbers. If you find that this is far more than that $5$% that you would get when creating plates randomly, then you have good reason to believe that lots of people got vanity plates. If, however, you find that this $P(N)$ is close to that $5$%, then that suggest not many people order vanity plates. So .. start counting! :)
You can solve this using Binomial Random Variable. Actually, you can continue my answer even if you are not familiar with the Binomial Random Variable.
Let $X$ be a random variable that takes value one if you see your desired number(2911) on the number plate.
$$\ \text{Total Number of Plates you see} = 100\text{days}\times 100\text{ plates/day}=10000\\ Pr(X=1) = {10000 \choose 10} \times ({ 1 \over 10000})^{10} \times ({ 9999 \over 10000})^{9990} \approx 1.0101\times 10^{-7} $$
We see that odds of seeing this number on ten different number plates is indeed very less.
Best Answer
$$p = (1/26) \cdot (1/26) \cdot (1/10)$$