Combining probabilities of independent attributes I've got what I think is a pretty simple stats question, but I haven't done this stuff in years.
Suppose I have 20 cars drive by an office everyday and someone is trying to id them based on 3 bits of data. Each "bit" is compromised of a finite set of values. I can id each car on its VIN number, but I'm trying to figure out how probable it is that someone else, without the VINs, could id any one of the 20 cars.
So, for example, I know the color is either red, green, or blue. I know the engine size is a 3.1L or a 4.0L. And I know the year is from 2001 to 2010. Each of these bits of data is independent of the others.
If I have 20 cars (and multiple combinations of these attributes match multiple cars):


*

*red, 3.1L, 2004

*green, 3.1L, 2006

*blue, 4.0L, 2010

*blue, 4.0L, 2001

*red, 3.1L, 2004

*red, 3.1L, 2004

*etc.


and each of these cars drives by everyday (assume they can see the engine size and year printed on the car :) ), can I figure out the odds that someone would be able to guess a given car based on this data?
(I know this analogy is contrived and falling apart, but the real data is too complex to go into.)
Would I just combine the probability of each bit of data?
Update
I'm asking for the likelihood of any combination given this data set of 20 observations. I'm not assuming assuming any prior knowledge about the likelihood of any combination, and in the data set there are clearly combinations that occur more frequently than others. Each car will only be observed once.
What I really want to know is, how likely is it that a set of attributes will have a 1-1 correlation to a car given the observations I have?
I hope that makes things clearer. Thanks.
 A: Am I correct in understanding that this question is equivalent to: "What is the probability that no two cars in 20 have identical descriptors?" 
There are $3 \times 2 \times 10=60$ ways to describe a car. This means that there are only 60 kinds of cars in our universe. Assuming that each way is equally probable, then this is a birthday-like problem with 20 randomly selected birthdays from a 60-day calendar. 
$$
P(20~unique~samples) = \frac{\binom{60}{20} \times 20!} {60^{20}} = .02789
$$
So you have a 97% chance that at least one pair of cars has exactly the same descriptors, so violating the requirement that all 20 cars be uniquely describable by the 3 values color, engine size and year.
It is a different problem to determine the distribution of the number of redundancies, which will be relatively few, so that most of the cars will be uniquely describable with the 3 factors.
Implementation note: since the numerator and denominator are large numbers, it is best to calculate the log of the ratio. I did this using R with the lchoose() and lfactorial() methods.

exp(lchoose(60,20)+lfactorial(20) - 20*log(60))

A: Case of uniform probabilities: The previous Answer by @PeterLeopold shows the exact combinatorial answer given that all 60 kinds of cars are equally likely. Here is a simulation in R that gets essentially the same
answer.
set.seed(2019); m = 10^6; n = 20
kinds = 1:60; pr = rep(1/60, 60)
d = replicate( m, n - length(unique(sample(kinds, n, repl=T, prob=pr))) )
mean(d == 0); prod(60:41)/60^20
 [1] 0.027511    # aprx P(no match) = 0.02789
 [1] 0.02789361
mean(d)
 [1] 2.87515      # aprx mean nr of matches

Here is a histogram of the simulated distribution of numbers of matches based on
a million iterations. A 'match' is any redundant ID; if cars A and B are of kind 1, and cars C and D are of kind 2, that is two matches; if cars A, B, and C are all of kind 1, that is also two matches.)

Simulation in a nonuniform case; The simulation above is based on the equally likely probability vector in which each kind
of car has probability $1/60$ of being chosen. There is no easy combinatorial method
for the case of kinds that are not equally likely. A simulation in such a case
can be done by replacing my vector pr by a vector of length 60 that gives the
actual probabilties of the respective kinds.
