My son is doing a math worksheet. He is practicing the concept of 'carrying' when doing sums. The sheet has 16 problems, each is summing two threedigit numbers. What struck me as interesting was that the creator of the problems made every problem have at least two carries and most have three.
What is the probability distribution for the number of carries in finding the sum of two randomly selected threedigit numbers?
Feel free to generalize.
(In reply to
solution by Charlie)
What's the probability that a set of 16 additions of 3digit numbers drawn at random will have only problems with at least two carries and furthermore that most have three carries?
The first condition has probability (9/25 + 583/3600)^16 ~= .0000303381063440692 or about 1 in 32961.8463545101171092383.
Given that all the additions have at least two carries what's the probability that a given addition has three carries? It's (583/3600)/(583/3500 + 9/25). That's about .307545065412671.
Then the probability that a majority (9 through 16) of the additions have three carries is:
.307545065412671 ^ 9 * .692454934587329 ^ 7 * C(16,9) +
.307545065412671 ^ 10 * .692454934587329 ^ 6 * C(16,10) +
.307545065412671 ^ 11 * .692454934587329 ^ 5 * C(16,11) +
.307545065412671 ^ 12 * .692454934587329 ^ 4 * C(16,12) +
.307545065412671 ^ 13 * .692454934587329 ^ 3 * C(16,13) +
.307545065412671 ^ 14 * .692454934587329 ^ 2 * C(16,14) +
.307545065412671 ^ 15 * .692454934587329 ^ 1 * C(16,15) +
.307545065412671 ^ 16 * .692454934587329 ^ 0 * C(16,16)
This sum is about .0301418726376864427. Dividing 32961.8463545101171092383 by this value gives a 1 in 1093556.68609978 probability of the observed types of practice additions coming up by chance.
Edited on January 13, 2017, 12:34 pm

Posted by Charlie
on 20170113 10:55:22 