What Is the Probability That Any Bucket Receives Exactly n Red Balls?

[paolopiace]

Greetings.
I would need help as follow up to the answers to my http://mathhelpboards.com/basic-probability-statistics-23/prob-red-ball-buckets-binomial-17282-post79735.html#post79735.

After having thrown R red balls over M buckets, the probability that exactly n red balls fall in one, specific, selected bucket is:

\(\displaystyle P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(1-\frac{1}{M}\right)^{R-n}\)

That applies to one, specific, selected bucket among M. That's why 1/M.

But I do not care to pick a specific bucket and look into it. Any bucket may receive n red balls and I need to look into all of them.

So, what is the probability to find n red balls into ANY bucket? If it sufficient to multiply the formula by M?

Thanks for your help!

 

[Arenholt]

Greetings.
I would need help as follow up to the answers to my http://mathhelpboards.com/basic-probability-statistics-23/prob-red-ball-buckets-binomial-17282-post79735.html#post79735.

After having thrown R red balls over M buckets, the probability that exactly n red balls fall in one, specific, selected bucket is:

\(\displaystyle P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(1-\frac{1}{M}\right)^{R-n}\)

That applies to one, specific, selected bucket among M. That's why 1/M.

But I do not care to pick a specific bucket and look into it. Any bucket may receive n red balls and I need to look into all of them.

So, what is the probability to find n red balls into ANY bucket? If it sufficient to multiply the formula by M?

Thanks for your help!

The probability calculated here is just for one bucket. For all buckets, $P_{total}=P_{m1}+P_{m2}+...P_{M}=M\times P$(since the probabilities for each bucket are identical). You can think about it intuitively as: if you don't focus on a particular bucket, the chances of you finding exactly "n-balls" in any bucket should increase $M-times$.

 

[paolopiace]

... You can think about it intuitively as: if you don't focus on a particular bucket, the chances of you finding exactly "n-balls" in any bucket should increase $M-times$.

Thank You! It's what I though but I needed confirmation.

 

[paolopiace]

The probability calculated here is just for one bucket. For all buckets, $P_{total}=P_{m1}+P_{m2}+...P_{M}=M\times P$(since the probabilities for each bucket are identical)...

Hummm... Are you sure?

So, if I have M=3 buckets and I throw R=2 red balls above them the prob. of finding n=1 red ball in one specific bucket is (from the binomial formula):
P(X=1) = 0.444

Based on what you said, M x P(X=1) = 1.333
This is not a probability.

 

[I like Serena]

Hummm... Are you sure?

So, if I have M=3 buckets and I throw R=2 red balls above them the prob. of finding n=1 red ball in one specific bucket is (from the binomial formula):
P(X=1) = 0.444

Based on what you said, M x P(X=1) = 1.333
This is not a probability.

Indeed. It's not true.
That formula only holds if the events are mutually exclusive... but they aren't.
The sum formula for events A and B that are not mutually exclusive is:
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$

Let's take a closer look at your example, and let's call $P(1)$ the probability we find $n$ balls in bucket $1$ of the $M=3$ buckets. And for instance $P(1 \cup 2)$ is the probability to find $n$ balls in bucket $1$ and/or in bucket $2$. And $P(1\cap 2)$ is the probability to find $n$ balls both in bucket $1$ and in bucket $2$. The very fact that the latter is possible means that the events are not mutually exclusive.
Then it boils down to:
$$
P_{total} = P(1 \cup 2 \cup 3) = P(1) + P(2) + P(3) - P(1\cap 2) - P(1\cap 3) - P(2\cap 3) + P(1\cap 2 \cap 3) \\
= \frac 49 + \frac 49 + \frac 49 - \frac 29 - \frac 29 - \frac 29 + 0 = \frac 69
$$

 

[paolopiace]

Indeed. It's not true.

Thank You!
Would you help me with the problem in my original post, here at the top?

I really need that probability.

 

[Arenholt]

Hummm... Are you sure?

So, if I have M=3 buckets and I throw R=2 red balls above them the prob. of finding n=1 red ball in one specific bucket is (from the binomial formula):
P(X=1) = 0.444

Based on what you said, M x P(X=1) = 1.333
This is not a probability.

For the given scenario, the formula indeed doesn't hold up. However, I had the doubt if there can only be one bucket with exactly n-balls or are multiple buckets each with n-balls allowed? Because if multiple buckets are allowed then we need to take cases to get the total probability.

 

[I like Serena]

Thank You!
Would you help me with the problem in my original post, here at the top?

I really need that probability.

Let's see... we can simplify the formula I just gave for your example a bit:
$$\begin{aligned}P_{total} &= \binom M1 P(1) - \binom M2 P(1 \text{ and } 2) + \binom M 3 P(1 \text{ and } 2 \text{ and } 3) - ... \\
&= \sum_{k=1}^{\min(M,\lfloor R/n \rfloor)} (-1)^{k-1}\binom Mk P(\text{exactly $n$ balls in buckets $1$ up to $k$})
\end{aligned}$$
And the probability to have exactly $n$ balls in buckets $1$ up to $k$ is:
$$\begin{aligned}P(\text{exactly $n$ balls in buckets $1$ up to $k$})
&= \underbrace{\binom Rn \binom {R-n}n ... \binom{R-(k-1)n}n}_{k\text{ factors}} \left(\frac 1M\right)^{kn} \left(1-\frac 1M\right)^{R-kn} \\
&= \prod_{i=0}^{k-1}\binom{R-in}n \left(\frac 1M\right)^{kn} \left(1-\frac 1M\right)^{R-kn}
\end{aligned}$$

 

[paolopiace]

Wow... I see. So, let's put the problem (the very 1st post of this thread) a bit differently.

Instead of asking "what is the probability to find n red balls into ANY bucket", I ask:

"What is the probability to find exactly n red balls into at least one bucket".

Here is the train of thoughts:

The prob. that one specific bucket has exactly n red balls is: P(X=n)

The prob. that one specific bucket has anything other than n red balls is: 1-P(X=n)

The prob. that all M buckets have anything other than n red balls is: (1-P(X=n))^M

The prob. that at least one bucket has exactly n red balls is: 1- (1-P(X=n))^M

Is this correct?

 

[I like Serena]

Wow... I see. So, let's put the problem (the very 1st post of this thread) a bit differently.

Instead of asking "what is the probability to find n red balls into ANY bucket", I ask:

"What is the probability to find exactly n red balls into at least one bucket".

Erm... what's the difference?
The formula I gave is the probability we have at least one bucket with $n$ balls.

Here is the train of thoughts:

The prob. that one specific bucket has exactly n red balls is: P(X=n)

The prob. that one specific bucket has anything other than n red balls is: 1-P(X=n)

The prob. that all M buckets have anything other than n red balls is: (1-P(X=n))^M

The prob. that at least one bucket has exactly n red balls is: 1- (1-P(X=n))^M

Is this correct?

Not quite.
The problem is that once we know a specific bucket does not have $n$ balls, the probabilities for the other buckets change.

 

[paolopiace]

...
And the probability to have exactly $n$ balls in buckets $1$ up to $k$ is: ...

Thank You. That's nice.
But it's not what the original problem asks. It states and asks:

After having thrown R balls over M buckets, the probability that exactly n<R balls fall in one specific bucket is:
$$P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(1-\frac{1}{M}\right)^{R-n}$$
That applies to one, specific, selected bucket among M. That's why 1/M.

But I do not care to pick a specific bucket and look into it. Any one bucket may receive n balls and I need to look into all of them.

So, what is the probability that at least one bucket, anyone among M, will contain n balls after having thrown R above the buckets?

 

[I like Serena]

Thank You. That's nice.
But it's not what the original problem asks. It states and asks:

After having thrown R balls over M buckets, the probability that exactly n<R balls fall in one specific bucket is:
$$P(X=n)={R \choose n}\left(\frac{1}{M}\right)^n\left(1-\frac{1}{M}\right)^{R-n}$$
That applies to one, specific, selected bucket among M. That's why 1/M.

But I do not care to pick a specific bucket and look into it. Any one bucket may receive n balls and I need to look into all of them.

So, what is the probability that at least one bucket, anyone among M, will contain n balls after having thrown R above the buckets?

That's the $P_{total}$ that I mentioned in that same http://mathhelpboards.com/basic-probability-statistics-23/binomial-probability-again-18881-post87939.html#post87939.
That is:
\[ P_{total} = P(\text{at least 1 bucket of the M buckets has n balls in it})\]