Quick Probability Question

[M.B.E.]

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If the item of interest is that an 11-digit phone number will contain only three digit values or fewer, but these can be ANY three digit values, we need to multiply the probability corresponding to any specific 3 digit
values by the number of such combinations of digit values. That does involve factorials:

P = (3/10)^11 * 10!/(7!3!) , approximately 0.000213

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Actually that's incorrect; it's too high because you are counting ESNs like 33433334343 eight times. That ESN would get counted for the three-digit sets {0,3,4}, {1,3,4}, {2,3,4}, {3,4,5}, {3,4,6} etc.

Also, ESNs like 66666666666 are being counted 36 times. It gets counted for the three digit sets {0,1,6}, {0,2,6} etc.

The correct answer is 0.000206128. That is the exact probability that a randomly chosen ESN contains three or fewer distinct digits (assuming all ESNs from 00000000000 through 99999999999 are equally likely).

The probability equals approximately 1/4851.