My Math Forum Probabilistic question

 November 5th, 2019, 12:49 AM #1 Newbie   Joined: Nov 2019 From: Romania Posts: 4 Thanks: 0 Probabilistic question Hi guys, I have a probability question that I really tried to solved, but seems to be to hard for my actual knowledge. I would kindly asked if you can help me with that. We have 985 white balls and 15 black balls, randomly distributed in the white balls. All 1000 balls are splited in groups of 10 balls, so we have 100 groups, each group having 10 balls. The questions is: which is the probability to have more than 1 black ball in any group of 10 balls? Can anyone please help me with this challenge? Many thanks in advance for your kind support! Lidia
 November 5th, 2019, 06:28 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 You can replace "white ball" with "nothing," so if I'm not mistaken, the question is the same as, "If we place each of 15 balls randomly in one of 100 boxes, what is the probability of any box containing more than one ball." There are two questions to be answered: 1) If we numbered the balls 1 - 15 (to keep probability in play), how many ways can we place the balls in the boxes? (Answer = 100^15) 2) Of these permutations, how many have 15 boxes with one ball each? Not an answer, but a start. Thanks from Lidia
 November 5th, 2019, 02:11 PM #3 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 The probability of all white balls $P_0=\frac{985\times 984\times ....976}{1000\times 999\times ...991}$. The probability of exactly one white ball $P_1=10\times \frac{15\times 985\times 984 ..\times .977}{1000\times 999\times ...991}$. Your answer $=1-P_0-P_1$. Thanks from Lidia
November 6th, 2019, 09:50 AM   #4
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Quote:
 Originally Posted by mathman The probability of all white balls $P_0=\frac{985\times 984\times ....976}{1000\times 999\times ...991}$. The probability of exactly one white ball $P_1=10\times \frac{15\times 985\times 984 ..\times .977}{1000\times 999\times ...991}$. Your answer $=1-P_0-P_1$.
Hi Mathman,

Thank you so much for your answer. The challange is to calculate which is the probability to have more than 1 black ball/set, let's say for example 2 balls.
Lidia

Last edited by Lidia; November 6th, 2019 at 09:53 AM.

November 6th, 2019, 09:52 AM   #5
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Quote:
 Originally Posted by DarnItJimImAnEngineer You can replace "white ball" with "nothing," so if I'm not mistaken, the question is the same as, "If we place each of 15 balls randomly in one of 100 boxes, what is the probability of any box containing more than one ball." There are two questions to be answered: 1) If we numbered the balls 1 - 15 (to keep probability in play), how many ways can we place the balls in the boxes? (Answer = 100^15) 2) Of these permutations, how many have 15 boxes with one ball each? Not an answer, but a start.
Hi DarnItJimImAnEngineer,

Thank you very much for your input. Really apreciate it!
Lidia

November 6th, 2019, 10:00 AM   #6
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Quote:
 Originally Posted by mathman The probability of all white balls $P_0=\frac{985\times 984\times ....976}{1000\times 999\times ...991}$. The probability of exactly one white ball $P_1=10\times \frac{15\times 985\times 984 ..\times .977}{1000\times 999\times ...991}$. Your answer $=1-P_0-P_1$.
I used your proposed solution and the result is = 0.127658
I also used other solution, by using HYPGEOMDIST function for excel (HYPGEOMDIST(1,10,15,1000). The result is different = 0.132039

 November 6th, 2019, 10:20 AM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1473 The probability of at least 1 group having 2 or more black balls is $1 - P[\text{no group has more than 1 black ball}] =1 - \dfrac{1}{N_t}$ where $N_t$ is the total number of distinguishable arrangements of 100 groups of 10 This is because there is only a single arrangement, 15 groups of (9,1) and 85 groups of (10,0), that have no groups with two or more black balls. The hard part, imo, is determining $N_t$ This is certainly related to the number of integer partitions of 15 using integers 0-10. But I haven't worked out exactly how it all fits together yet. Maybe you can.
 November 6th, 2019, 06:22 PM #8 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 Side note, if I had the time currently to work on this, I would probably start with a smaller problem, where I could easily count possibilities. 3 black balls in 3 boxes. Then 3 black balls in 4 boxes, 5 boxes, 6 boxes, …, until I found a pattern that looked similar to probability formulas. Then check against a 4 black ball case and maybe a 5 black ball case. Thanks from romsek
November 6th, 2019, 07:04 PM   #9
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Quote:
 Originally Posted by DarnItJimImAnEngineer Side note, if I had the time currently to work on this, I would probably start with a smaller problem, where I could easily count possibilities. 3 black balls in 3 boxes. Then 3 black balls in 4 boxes, 5 boxes, 6 boxes, …, until I found a pattern that looked similar to probability formulas. Then check against a 4 black ball case and maybe a 5 black ball case.
always an excellent idea

 November 6th, 2019, 07:59 PM #10 Senior Member     Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1473 OP has apparently obtained an answer as noted on another forum.

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