Probability question

[Lidia]

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

 

[Deleted member 4993]

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

Please share your work/thoughts about this assignment.

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

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[pka]

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?

This is a rather complicated question that could involve Stirling numbers.
I find your last statement a little confusing. Any group could have anywhere from zero to fifteen black balls.
So please tell exact meaning of "the probability to have more than 1 black ball in any group of 10 balls"?

 

[Lidia]

This is a rather complicated question that could involve Stirling numbers.
I find your last statement a little confusing. Any group could have anywhere from zero to fifteen black balls.
So please tell exact meaning of "the probability to have more than 1 black ball in any group of 10 balls"?

Hi Pka,

Thank you so much for your reply! Yes, let me give you more detailes about this subject. I will explain better with an operational example.
Let's suppose we produce 1000 bottles and we deliver them to the customer packed in carton pack, having 10 bottles/pack. So, totaly we have 100 packs with 10 bottle/pack. Let's suppose that we know from the experience that we have 15 bottles with defects generated at 1000 bottles produced. The question is, which is the probability to have more than one bottle with deffect in one pack. In other words, which is the probability to have 85 packs with bottle witout deffects, 14 packs with bottles having one bottle with defect/pack and 1 pack having 2 bottles with deffects.
Start from here, which is the probability to have 2 packs having 2 bottles with deffects/each pack, 3 pack having 2 bottles with deffects/each pack.
Many thanks in advance for your kind reply!
Lidia

 

[Steven G]

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

If I am understanding this correctly I would find the probability of all groups of 10 balls having 0 or 1 ball. That problem removes the white balls completely. Think of this: You have 100 containers to place the 15 black balls into and you want to know in how many cays can you do this if you can't place two or more balls into any one container. That is my take on this problem.

 

[pka]

If I am understanding this correctly I would find the probability of all groups of 10 balls having 0 or 1 ball. That problem removes the white balls completely. Think of this: You have 100 containers to place the 15 black balls into and you want to know in how many cays can you do this if you can't place two or more balls into any one container. That is my take on this problem.

Jomo, the post says that all 1000 balls are put in one hundred groups of ten each. So that has to include the fifteen black balls among the 985 black balls.

 

[Steven G]

Jomo, the post says that all 1000 balls are put in one hundred groups of ten each. So that has to include the fifteen black balls among the 985 black balls.

I was thinking that it is reasonable that the 15 black balls could have been put into one of the 100 piles first, then the 985 white balls and then my idea works?

 

[pka]

I was thinking that it is reasonable that the 15 black balls could have been put into one of the 100 piles first, then the 985 white balls and then my idea works?

I doubt that. But can you explain the \meaning of """the probability to have more than 1 black ball in any group of 10 balls"?

 

[Steven G]

I doubt that. But can you explain the \meaning of """the probability to have more than 1 black ball in any group of 10 balls"?

Sure. You have 1000 balls which you will place into 100 containers with exactly 10 balls in each. 15 of the balls will be black and 985 balls will be white. Some containers will end up with no black balls, some containers may end up with 1 black ball, .... The question is asking what is the chance that a container has more than 1 black ball. An example of this is if container 1 has zero black balls, container 2 has three black balls, container 3 has two black balls, .....

 

[Lidia]

Hi all,

Thank you so mych for your precious inputs on this subject. I found a solution on this challange by using HYPGEOMDIST function from excel. Works great!
Many thanks again!
Lidia

 

[Steven G]

Please show us your solution so others can benefit from it. Thanks!

 

[Lidia]

Please show us your solution so others can benefit from it. Thanks!

Hi Jomo,
Yes, I didn't explain my solution. I apologize...

As I said, I used HYPGEOMDIST function from excel as following:

P =HYPGEOMDIST(1,10,15,1000), where:

1 = Sample_s The number of successes in the sample.
10 = Number_sample The size of the sample.
15 = Population_s The number of successes in the population.
1000 = Number_pop The population size.

Thank you so much again to you and to all team for the knowledge share with me!
Lidia