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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 $P$ The max sum will be 9, so you have to repeat this for 6-9. It will probably be easier to calculate $P$, add this to the $P,~P$ you just calculated, and subtract that all from 1 to obtain $P[6+]$ June 16th, 2019, 06:36 PM   #5
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Quote:
 Originally Posted by romsek It will probably be easier to calculate $P$, add this to the $P,~P$ 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. Tags balls, drawn, probability, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post EvanJ Probability and Statistics 4 June 26th, 2017 05:33 PM Skyer Algebra 3 January 6th, 2014 11:26 AM rayman Probability and Statistics 1 May 21st, 2013 10:21 AM baxy7 Advanced Statistics 2 August 6th, 2011 01:46 PM sara_18 Probability and Statistics 2 April 3rd, 2010 08:01 PM

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