Suppose that in a certain state, all automobile license plates have three letters followed by four numerical digits. Two letters may never be used (I, O). If the license plates are randomly generated, what is the probability that a license plate will contain “UMW”?

Question

anonymous · Answer

@satellite73 if you could show me how to do this-- that would be great...

kropot72 · Answer

There are 24 letters from which to choose. Assuming that repetitions are allowed, the number of permutations of three of the 24 letters is 24^3. The probability a particular permutation will appear on as license plate is given by:
$$\large \frac{1}{24^{3}}$$

anonymous · Answer

@kropot72 would we then multiplly that by 3?

anonymous · Answer

woops. i guess it's just what you wrote there. sorry didn't recognize the ^3

kropot72 · Answer

Assuming that repetitions of letters are allowed, there are 24 choices for the first letter, 24 choices for the second letter and 24 choices for the third letter. This gives a total of 24^3 permutations. The probability of a particular permutation being randomly selected is simply the reciprocal of 24^3.
Why do you think this should be multiplied by three?