Scanning a number from left to right, if the next digit is greater in value than the preceeding digit, we say that the number is strictly increasing: 1589, 3679, and 1348 are examples of 4-digit numbers with this property.
Given a certain number is strictly increasing, what is the probability that it contains
a. exactly 6 digits?
b. 6 digits or more?
Assuming that what is wanted is the probability that from among all such numbers, one chosen at random with equal probability, will be exactly (a) 6 digits long or (b) at least 6 digits long. We also assume that there is no leading zero, and therefore no zero at all.
There are 2^9 - 1 = 511 such numbers.
digits number
1 9
2 C(9,2)=36
3 C(9,3)=84
4 C(9,4)=126
5 C(9,5)=126
6 C(9,6)=84
7 C(9,7)=36
8 C(9,8)=9
9 C(9,9)=1
Part a: 84/511 = 12/73 ~= .1643835616438356
Part b: 130/511 ~= .25440313111546
If, by the definition given, single-digit numbers are not eligible, then subtract 9 from each of the denominators of 511, and reduce from there: 84/502 = 42/251 ~= .1673306772908366 and 130/502 = 65/251 ~= .2589641434262948.
The above hinges on each of the 511 or 502 such numbers being equally likely to be chosen by the randomization. Numbers randomly chosen from the real world tend to follow Benford's Law, as well as constraints on size as determined by the type of data being considered.
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Posted by Charlie
on 2016-01-26 11:37:16 |
simple understanding, and randomization procedures
