Re: Question about License Plates

On Jun 27, 8:19 pm, "Stratocaster" <sto...@xxxxxxxxxxx> wrote:
<rbaul...@xxxxxxxxxxx> wrote in message

news:1182974023.791880.213460@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



Hello -

I've got a statistical question about license plate viewings for
anyone willing to field it. In New York State, license plates take
the format ABC-XXXX, where ABC is any alpha letter, and XXXX is a
number from 0000 to 9999.

Without knowing the specifics in terms of the exact number of cars on
the road at any given time --

If one were to pick any three digit number, is there any way to
determine a very estimated probability of seeing that number in a
license plate 3 times in a 3 hour viewing window, where less than 50
cars are randomly viewed? For example, if the number was 554, one
could see ABC-0554..ABC9554, or ABC-5540..ABC-5549 (where again ABC is
an any alpha combo). Would the probability be essentially near 0?

Thank you,

RB

If you allow some (big and unreasonable) simplifications, there can be an
easy answer.

1) ABC is any alpha combo (presumably three letters?)-Ok, this one is not
so BIG
2) Assume that all possible liscense plates exists (i.e. all ABC-XXXX
combos)
3. People drive around to random locations, from any location in
NY-state... (i.e. you are as likely to see your neighbors car as you are a
car from the other side of the state).
4). There do not exist personalized liscense plates.
5). Nobody from out of state ever drives by.

First, there are (26)^3 possible alpha combos.
Second, given a 3 digit number XXX, there are (10)^2 ways of ordering these
3 numbers contingently.

Now, there exist [(26)^3]*[10,000] unique liscense plates.
And there exist [(26)^3]*[100] possible liscense plates which contain the
randomly chosen 3 digit number.

So the probability that any randomly chosen car possesses a liscense plate
with these 3 numbers is:
[(26)^3]*[100]/[(26)^3]*[10,000]

1 out of 100

At this point I am unsure how to continue. Do you mean that 50 cars are
typically viewed in a three hour time period or do you mean there are 50
cars within the course of an entire day???

If the former, try looking up the binomial distribution. Your probability
of success would be .01, your probability of failure would be .99. Your
sample size would be 50, the number of successes you are testing for would
be 3. If the latter, more simplifications may be necessary (i.e. the road
has the same traffic density for any given hour, night or day).

Try it out, see how far you get.


Thank you for the reply.

50 cars would be viewed in the three hour time period where the
subject is riding in the car and randomly viewing cars.

Unfortunately, I'm not a statistician myself, so I don't know how to
look up the binomial distribution.

Could the probability of all 3 events happening, given the constraints
you list, be determined by simply multiplying (1/100)^3, or 1 in 1
million chance?

Thanks,

RB

.

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