My Math Forum Probabilities of the Sum of Four Digits

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 January 6th, 2017, 05:03 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 464 Thanks: 70 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, 06:23 PM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 464 Thanks: 70 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, 06:30 PM   #3
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and divide these all by 10000 for the probability
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