Question 1028988: If you were to make up a random 13-digit number, what is the probability that it could pass as a legitimate ISBN based on the check digit?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
If you were to make up a random 13-digit number, what is the probability that it could pass as a legitimate ISBN based on the check digit?
Solution:
First we have to know how the check digit works for an ISBN number.
Say the ISBN13 number is
9780486462370
The last digit is the check digit, and will not enter into the following calculations.
Step 1:
calculate the sum of all the even and odd position digits. Do not add the check digit, which is counted as the first odd position.
sum of odd positions: 3+6+6+4+8+9=36=S1
sum of even positions: 7+2+4+8+0+7=28=S2
Step 2:
calculate the checksum = S1+3S2=36+3*28=120
Step 3:
Find the remainder of the checksum divided by 10.
120/10=12 remainder 0.
Step 4:
Subtract the remainder by 10 to find the check digit.
If the check digit is 10, take it as zero.
10-0=10 therefore the check digit is zero, which is exactly what's in the last digit.
To proceed to calculate the probability, we first examine the calculation of the check digit.
It relates to the sum of the random odd digits and 3 times the sum of even digits. Knowing that all digits are random, we can almost say that the check digit is also random, i.e. with a probability of 1/10.
However, we have to confirm if the multiplication by 3 (of the even digits) yield a digit evenly distribution in the [0-9] range.
Check the final digit of the multiplication by three:
0 0
1 3
2 6
3 9
4 2
5 5
6 8
7 1
8 4
9 7
Thus, multiplication by three merely changes the digit in an evenly distributed over the range of [0-9], so therefore remains random.
Since the check digit is randomly distributed over the range [0-9], we can say that a randomly created 13-digit number will qualify as an ISBN13 number with a probability of 1/10.
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