My Math Forum Probabilities of the Sum of Four Digits

 Probability and Statistics Basic Probability and Statistics Math Forum

 January 6th, 2017, 04:03 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82 Probabilities of the Sum of Four Digits Four digits from 0 to 9 are added, with all the digits equally likely to be selected at any time. What are these probabilities? P(Sum = 0) = 1/10,000 P(Sum = 1) = 4/10,000 P(Sum = 2) = ? P(Sum = 3) = ? P(Sum = 4) = ? P(Sum = 5) = ? P(Sum = 6) = ? P(Sum > 6) = ? The sum of the numerators I didn't solve for has to be 9,995.
 January 6th, 2017, 05:23 PM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82 I realized that the sum of two digits has 19 possibilities, which are the numbers from 0 to 18. Then I made a spreadsheet with 19^2 = 361 rows, which didn't take long. I didn't want to solve the problem by writing out 10,000 rows, and I didn't have to. I figured out the answers on my own: P(Sum = 0) = 1/10,000 P(Sum = 1) = 4/10,000 P(Sum = 2) = 10/10,000 P(Sum = 3) = 20/10,000 P(Sum = 4) = 35/10,000 P(Sum = 5) = 56/10,000 P(Sum = 6) = 84/10,000 P(Sum > 6) = 9,790/10,000
January 6th, 2017, 05:30 PM   #3
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,980
Thanks: 1027

and divide these all by 10000 for the probability
Attached Images
 Clipboard01.jpg (19.3 KB, 10 views)

 Tags digits, probabilities, sum

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Monox D. I-Fly Probability and Statistics 2 October 17th, 2016 08:00 PM RifkiNada Algebra 2 November 24th, 2012 03:32 AM CarpeDiem Elementary Math 7 July 13th, 2012 07:35 PM safyras Algebra 23 March 22nd, 2011 09:59 PM Zerazar Advanced Statistics 3 April 15th, 2010 08:17 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top