probability of randomly chosen items being the same in 3 pots

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jugsy67

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Hi I have a probability poser.

if I have 3 jars, let's call them, Jar 1, Jar 2 and Jar 3, and each jar contains 1000 balls numbered 1 to 1000.

From Jar 1 I take 160 balls, as randomly as I can, and put them into a new Jar called Jar A.

I then repeat this task for Jar 2 into Jar B, and finally Jar 3 into Jar C.

So I now end up with 3 Jars, A, B and C, each containing a random collection of 160 balls, which can be any number from 1 to 1000.

Q1: What is the probability that Jars A, B and C all have ball the number 4 ball in them?
Q2: What is the probability that Jars A, B and C all have ball the number 4 and 20 balls in them?
Q3: What is the probability that Jars A, B and C all have the same numbered balls in them?
Q4: What is the formulae to work out the probability for any number balls, e.g. 6 or 20 balls, being the same in all 3 jars or just 2 of the jars

cheers

Julian
 
jugsy67 said:
Hi I have a probability poser.

if I have 3 jars, let's call them, Jar 1, Jar 2 and Jar 3, and each jar contains 1000 balls numbered 1 to 1000.

From Jar 1 I take 160 balls, as randomly as I can, and put them into a new Jar called Jar A.

I then repeat this task for Jar 2 into Jar B, and finally Jar 3 into Jar C.

So I now end up with 3 Jars, A, B and C, each containing a random collection of 160 balls, which can be any number from 1 to 1000.

Q1: What is the probability that Jars A, B and C all have ball the number 4 ball in them?
Q2: What is the probability that Jars A, B and C all have ball the number 4 and 20 balls in them?
Q3: What is the probability that Jars A, B and C all have the same numbered balls in them?
Q4: What is the formulae to work out the probability for any number balls, e.g. 6 or 20 balls, being the same in all 3 jars or just 2 of the jars

cheers

Julian
Click to expand...
First work with smaller numbers.

Suppose you put only one ball each in jars A, B & C. What is the probability that Jars A, B and C all have ball the number 4 ball in them?

Suppose you put only two balls each in jars A, B & C. What is the probability that Jars A, B and C all have ball the number 4 ball in them?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
This is a great problem. I can't wait to see how you tried working it out so that I can help you get to the correct answer. BTW, did you read the posting guidelines?
 
Is the event Jar A and Jar B to have ball #4 in them independent or dependent events?
What is the probability that ball 4 ends up in Jar A?
 
jugsy67 said:
if I have 3 jars, let's call them, Jar 1, Jar 2 and Jar 3, and each jar contains 1000 balls numbered 1 to 1000.
From Jar 1 I take 160 balls, as randomly as I can, and put them into a new Jar called Jar A.
I then repeat this task for Jar 2 into Jar B, and finally Jar 3 into Jar C.
So I now end up with 3 Jars, A, B and C, each containing a random collection of 160 balls, which can be any number from 1 to 1000.
Q1: What is the probability that Jars A, B and C all have ball the number 4 ball in them?
Click to expand...
These questions involve huge numbers. There SEE HERE ways to select \(160\) balls from \(1000\).
That is a 190 digit number. Of all of those, there SEE HERE that contain the ball numbered four. Can you explain that?
So the probability that Jar A contains the ball numbered four is SEE HERE
See what you can do with that much. You must post some work. Try Q2.
 
Hi Guys. for the one ball problem the chances that all jars had the number 4 ball I worked out was 1 in 1000^3 so 1 in 1 billion (US billion), however, for 2 balls the thing got more complex, as the second ball to extracted is now 1 in 999 chance of it being in Jar A. So that's why I gave up and hoped there was a formulae. If you wonder why this important to me it is because I am a microbiologist working on the human gut, and everyone's colon has about 160 bacterial species in it - selected from about 1000 bacterial species. So I was wondering what is the probability, that starting with 1000 bacterial species, that 2 people have the same collection of 160 bacterial species in their colon.
 
Q1: First find the probability of choosing a 4 into a given jar.

[MATH]P[\text{a 4 is chosen in 160 out of 1000}] = \dfrac{\dbinom{1}{1}\dbinom{999}{159}}{\dbinom{1000}{160}}[/MATH]
As pka noted these are enormous numbers... However they mostly cancel out, i.e.

[MATH]\dfrac{\dbinom{1}{1}\dbinom{999}{159}}{\dbinom{1000}{160}} = \dfrac{999!}{159!840!}\dfrac{160!840!}{1000!} = \dfrac{160}{1000} = \dfrac{4}{25} [/MATH]
Now as these separate jar picking events are independent we simply cube this single probability to get the probability that all 3
jars contain a 4.

[MATH]P[\text{all 3 jars contain a 4}] = \left(\dfrac{4}{25}\right)^3 = \dfrac{64}{625} \approx 0.1[/MATH]
Q2: Is slightly more interesting. 2 items in the select group now.

[MATH]P[\text{a jar contains both 4 and 20}] = \dfrac{\dbinom{2}{2}\dbinom{998}{158}}{\dbinom{1000}{160}} = \\~\\ \dfrac{998!}{158!840!}\dfrac{160!840!}{1000!} = \dfrac{160\cdot 159}{1000\cdot 999} = \dfrac{212}{8325} [/MATH]
and again we cube this to get the probability that all 3 jars contain a 4 and a 20

[MATH]P[\text{all 3 jars contains both 4 and 20}] = \left(\dfrac{212}{8325}\right)^3 = \dfrac{9528128}{576969328125} \approx 1.65141 \times 10^{-5}[/MATH]
Q3: Following above

[MATH]P[\text{all 3 jars have the same 160 balls}] = \dbinom{1000}{160} \cdot \left(\dfrac{\dbinom{160}{160}\dbinom{840}{0}}{\dbinom{1000}{160}}\right)^3 = \dbinom{1000}{160}^{-2} \approx 1.1 \times 10^{-379} [/MATH]
Q4:

[MATH]P[\text{k matches in all 3 jars}] = \dbinom{1000}{k} \left(\dfrac{\dbinom{1000-k}{160-k}}{\dbinom{1000}{160}}\right)^3[/MATH]
I'll let you figure out the probability of k matches in 2 jars.
 
Romsek said:
Q1: First find the probability of choosing a 4 into a given jar.

[MATH]P[\text{a 4 is chosen in 160 out of 1000}] = \dfrac{\dbinom{1}{1}\dbinom{999}{159}}{\dbinom{1000}{160}}[/MATH]
As pka noted these are enormous numbers... However they mostly cancel out, i.e.

[MATH]\dfrac{\dbinom{1}{1}\dbinom{999}{159}}{\dbinom{1000}{160}} = \dfrac{999!}{159!840!}\dfrac{160!840!}{1000!} = \dfrac{160}{1000} = \dfrac{4}{25} [/MATH]
Now as these separate jar picking events are independent we simply cube this single probability to get the probability that all 3
jars contain a 4.

[MATH]P[\text{all 3 jars contain a 4}] = \left(\dfrac{4}{25}\right)^3 = \dfrac{64}{625} \approx 0.1[/MATH]
Q2: Is slightly more interesting. 2 items in the select group now.

[MATH]P[\text{a jar contains both 4 and 20}] = \dfrac{\dbinom{2}{2}\dbinom{998}{158}}{\dbinom{1000}{160}} = \\~\\ \dfrac{998!}{158!840!}\dfrac{160!840!}{1000!} = \dfrac{160\cdot 159}{1000\cdot 999} = \dfrac{212}{8325} [/MATH]
and again we cube this to get the probability that all 3 jars contain a 4 and a 20

[MATH]P[\text{all 3 jars contains both 4 and 20}] = \left(\dfrac{212}{8325}\right)^3 = \dfrac{9528128}{576969328125} \approx 1.65141 \times 10^{-5}[/MATH]
Q3: Following above

[MATH]P[\text{all 3 jars have the same 160 balls}] = \dbinom{1000}{160} \cdot \left(\dfrac{\dbinom{160}{160}\dbinom{840}{0}}{\dbinom{1000}{160}}\right)^3 = \dbinom{1000}{160}^{-2} \approx 1.1 \times 10^{-379} [/MATH]
Q4:

[MATH]P[\text{k matches in all 3 jars}] = \dbinom{1000}{k} \left(\dfrac{\dbinom{1000-k}{160-k}}{\dbinom{1000}{160}}\right)^3[/MATH]
I'll let you figure out the probability of k matches in 2 jars.
Click to expand...
many thanks
 
guys i am humbled by your help and offering your time to help me solves this - as you can see a microbiologist I am clueless when it comes to probability :)
 
Jomo said:
This is a great problem. I can't wait to see how you tried working it out so that I can help you get to the correct answer. BTW, did you read the posting guidelines?
Click to expand...
yes I did and didn't show my working because I had no idea where to start - as you can see my working was so way off target.
 
jugsy67 said:
many thanks
Click to expand...
may I ask why for choosing Ball Number 4 from Jar 1, which contains 1000 balls and putting it into Jar that is not 1 in 1000 chance? If it was rolling 3 sixes on a 6 sides dice with 3 rolls of the dice the probability would be 1 in 216. Why is putting one ball -called number 4 into Jar A a 4 in 25 chance?
 
jugsy67 said:
may I ask why for choosing Ball Number 4 from Jar 1, which contains 1000 balls and putting it into Jar that is not 1 in 1000 chance? If it was rolling 3 sixes on a 6 sides dice with 3 rolls of the dice the probability would be 1 in 216. Why is putting one ball -called number 4 into Jar A a 4 in 25 chance?
Click to expand...
You are choosing 160 balls - not 1.
 
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