My Math Forum Probability of Sum of Balls Drawn

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 June 16th, 2019, 08:25 AM #1 Newbie   Joined: Jun 2019 From: UK Posts: 2 Thanks: 0 Probability of Sum of Balls Drawn Hi, I'm sure this is simpler than my brain is making it out to be but I need help... So, we've got a bag of 20 balls and the user draws 3 balls out. I'm trying to work out the probability of the total value of those 3 balls being 4-5 or 6 and over. The balls have the following values and quantities: A - 1 (12 balls) B - 2 (6 balls) C - 3 (2 balls) It's driving me mad so if someone could help it'd be great - thanks!
 June 16th, 2019, 10:49 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 It's probably easiest to calculate the probability of 3 and subtract that from 1. There's only 1 combo that sums to 3 (3 is the minimum sum), and that's 3 1's. $P[\text{3 1's}] = \dfrac{12}{20}\dfrac{11}{19}\dfrac{10}{18} = \dfrac{11}{57}\\ P[\text{sum of 4 or more}] = 1 -P[\text{3 1's}] = 1 - \dfrac{11}{57}=\dfrac{46}{57}$
 June 16th, 2019, 11:33 AM #3 Newbie   Joined: Jun 2019 From: UK Posts: 2 Thanks: 0 Hey, thanks for the reply! I should have been more specific maybe - I want to specifically calculate the probability of the sum of the three balls being 4 or 5 and then I want to calculate the probability of it being 6 or more.
 June 16th, 2019, 12:04 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 I see. Well what you need to do is find the 3 element partitions of each of these numbers, 4, 5, 6+ And then calculate the probability of each member of the partition. For example for n=4. $4 = (2,1,0)$, i.e. 2 1's, 1 2, and 0 3's. We don't care about order. This is the only way to sum to 4 using 3 draws. $P[(2,1,0)] =\dfrac{\dbinom{12}{2}\dbinom{6}{1}\dbinom{2}{0}}{ \dbinom{20}{3}} = \dfrac{66 \cdot 6}{1140}=\dfrac{11}{190}$ Similarly the partitions of 5 are $5 = (2,0,1),(1,2,0)$ Just apply the same method for calculating the probability of each of these partitions and sum them to obtain $P[5]$ The max sum will be 9, so you have to repeat this for 6-9. It will probably be easier to calculate $P[3]$, add this to the $P[4],~P[5]$ you just calculated, and subtract that all from 1 to obtain $P[6+]$
June 16th, 2019, 06:36 PM   #5
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 Originally Posted by romsek It will probably be easier to calculate $P[3]$, add this to the $P[4],~P[5]$ you just calculated, and subtract that all from 1 to obtain $P[6+]$
If you are taking a timed test, you should calculate the probability of the sum being 3, 4, and 5 and do 1 minus that to get the probability of 6+. If you are trying to learn, you could calculate the probability of every sum and check that it adds up to 1 to see if you are doing it correctly. That's because you could get a similar problem where it isn't quicker to get what you need by calculating 1 minus other probabilities. I recently used the binomial probability formula to calculate the probability of every amount of successes from 0 to 10 of 10 trials, and I confirmed that I had the correct sum.

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