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?
(In reply to
Counting, Counting (spoiler) by Steve Herman)
Actually if you allow decimals you can have a leading zero after the decimal point. 12 can be 12 or 1.2 or .12 or .012 (four possibilities.) A six digit number then has 8 possibilities.
I see no reason to stick to 2 or more digit numbers but I won't include zero. So my count, including decimals:
C(9,1)*3=27
C(9,2)*4=144
C(9,3)*5=420
C(9,4)*6=756
C(9,5)*7=882
C(9,6)*8=672
C(9,7)*9=324
C(9,8)*10=90
C(9,9)*11=11
For a total of 3326
So part a) = 672/3326 or about .202
part b) = 1097/3326 or about .330
There is also an implication that the number is chosen with uniform probability from the set of all 3326 such numbers and not as the end result of some process.
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Posted by Jer
on 2016-01-26 10:27:10 |
re: Counting, Counting (spoiler) - decimals
