Revising probability using Bayes' rule Let's say $100$ numbers is picked from a set of numbers one by one with replacement, which is between $000$ to $999$. I'm required to give my subjective probability before the experiment and revise (or update) the my subjective probability based on the results obtained. I would like to know if I'm doing it correctly or is it logical to use Bayes' rule for revision: (I'am so confused with my answers)
Since there are $^{10}P_{3}=(10)(9)(8)=720$ 3-different-digits number (e.g. $012,256$) from $000-999$, my prior probability, $P(H)=\frac{720}{1000}=0.72$
Based on the results obtained, $80$ out of $100$ numbers are $3$-different-digits number. So the probability based on the data, $P(E)=\frac{80}{100}=0.80$
According to Bayes' rule, (from Wikipedia)
$$P(H|E)=\frac{P(E|H)\cdot P(H)}{P(E)}$$
where $P(H|E)$ is the posterior probability, which will also be my revised probability for this question
$P(E|H)$, or the likelihood, is calculated as follows:
$$P(E|H)=\binom{100}{80}(0.72)^{80}(0.28)^{20}=0.01815146392$$
Hence, 
$$P(H|E)=\frac{(0.01815146392)(0.72)}{0.80}=0.01633631753$$
Is there something wrong with my method? Because my revised probability is so small compared to my probability given before the experiment ...... 
 A: You are mixing up two different "levels" of probability. You have an unknown distribution for the bag of numbers, and let $H$ denote the proportion of mixed digit numbers in the bag.  Your prior probability should be a distribution over the possibilities for $H$ (which is some number between $0$ and $1$, you don't know how many numbers are in the bag even!). So, you assumed a prior that suggests with certainty that the proportion of mixed digit numbers in the bag is $0.72$, i.e. your prior distribution is $P(H=0.72) = 1$ and $P(H\neq 0.72) = 0$. 
This doesn't seem like a good starting guess (and it looks like if you applied the update rule anyway, you get the same distribution back, deterministically 0.72). Maybe you should start with $H$ being the uniform distribution on $[0,1]$?
Then the update rule will update the distribution over $H$. With the given observation you'd expect the new distribution of $H$ to become more centered towards $0.80$ after this one update?  In this case you'll have to compute conditional densities.
Alternatively, if you want to work with a discrete distribution, you could assume a prior that $H$ can only be one of $\{0,0.25,0.5,0.75,1\}$ with equal probability (also probably not a good prior, as in the limit the update will center on one of these values, even though it's possible the bag has a proportion other than these ones), and after one update at least $0$ and $1$ will be eliminated. The point of the question I guess is to get a feel for how updates work.
