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June 16th, 2019, 08:25 AM   #1
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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!
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June 16th, 2019, 10:49 AM   #2
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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.

$P[\text{3 1's}] = \dfrac{12}{20}\dfrac{11}{19}\dfrac{10}{18} = \dfrac{11}{57}\\
P[\text{sum of 4 or more}] = 1 -P[\text{3 1's}] = 1 - \dfrac{11}{57}=\dfrac{46}{57}$
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June 16th, 2019, 11:33 AM   #3
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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.
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June 16th, 2019, 12:04 PM   #4
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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 $P[6+]$
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June 16th, 2019, 06:36 PM   #5
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Quote:
Originally Posted by romsek View Post
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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