Advanced (Or simple?) balls in a jar probability

• csb
In summary: B20 (5 unique balls)jar 5 : B3, B10, B2, B14, B16 (5 unique balls)and so on...My question is:What is the probability of observing one particular ball type in a jar?
csb
If I have 8 jars, each jar contains 5 unique ball types. However, I know that I have 20 unique ball types out there. So, I have balls labelled from B1, B2, B3, ...B20 to put into 5 jars.

Let's say

Jar 1 : B1, B3, B4, B7, B12 (5 unique balls)

jar 2 : B9, B7, B11, B15, B18 (5 unique balls)

jar 3 : B20, B1, B5, B17, B13 (5 unique balls)

jar 4 : B4, B5, B6, B3, B17 (5 unique balls)

jar 5 : B3, B10, B2, B14, B16 (5 unique balls)

and so on...

My question :

1. What is the probability to observe one particular ball type in a jar ?
2. What is the probability to observe one particular ball type in at least n jars ?
Is there a well-known probability distribution or combinatorial method that might relate to this ?

For question 1, can I say that the probability is 1/5 ? Or is it 5C1/20C5 ? Or should I use more complicated thing like Stirling number of the second kind for this problem ? Totally not sure...

For question 2, once I know the probability from question 1, I can use binomial distribution, right ?

Thanks

The answer is simple, but it is not 1/5. What if there were 5 different ball types? Or 6?

I think the answer to number 1 is 1/20. the p of the ball being in a jar is 5/20 times p of picking the ball 1/5. assuming independence.

for number two you need the answer to number one times the p that the ball is in another jar*you pick that jar*you pick that ball

but you also need to accoount for the p that the ball is in the space again but in the same jar that you already picked, this can make my answer for number 1 incorrect so I would say it is not an easy problem. 5C1/20C5 is not right it is way to small.

There 20 unique ball types in total. If you take one jar, how many unique ball types are there in it? And how many unique ball types are there in total? So, assuming that any ball type is equally likely to be in the jar, what is the probability that any given ball type is one of the types that is in the jar?

The probability that any ball type is one of the ball types in the jar is 1 because they are all a ball type.

Is the question what is the probability that a particular ball is in at least one jar?

I thought it was; if you distribute 20 unique balls uniformly into 5 distinct jars, 5 balls in each jar, each ball put in a jar independently with p=1/20, what is the probability that you will choose one particular ball out of a jar?

I'm assuming question 1 is equivalent to: I randomly select 5 out of the 20 available types. What is the chance that type 1 is one of the 5 selected?

The answer to that question is 1 - (19C5/20C5), which is 1/4. That's because there are 20C5 ways to select the 5 types, and 19C5 ways to select the 5 types without choosing type 1.

Na because that's without replacement. you can repeat balls so it is 1-19^5/20^5 = .2262

davidmoore63@y said:
I'm assuming question 1 is equivalent to: I randomly select 5 out of the 20 available types. What is the chance that type 1 is one of the 5 selected?

The answer to that question is 1 - (19C5/20C5), which is 1/4. That's because there are 20C5 ways to select the 5 types, and 19C5 ways to select the 5 types without choosing type 1.
That is the right answer, but it is more easily arrived at by considering that there are 5 mutually exclusive 1/20 chances for a given ball type to be in a given jar: the answer is 5 x 1/20 = 1/4.

MrAnchovy said:
That is the right answer, but it is more easily arrived at by considering that there are 5 mutually exclusive 1/20 chances for a given ball type to be in a given jar: the answer is 5 x 1/20 = 1/4.

19C5/20C5 is not correct because it doesn't consider cases such as

B2 B2 B3 B3 B4
B2 B2 B2 B2 B2
or
B2 B3 B4 B5 B5

csb said:
each jar contains 5 unique ball types

O god dam

So binomial with p=.25

csb said:
If I have 8 jars, each jar contains 5 unique ball types. However, I know that I have 20 unique ball types out there. So, I have balls labelled from B1, B2, B3, ...B20 to put into 5 jars.

Let's say

Jar 1 : B1, B3, B4, B7, B12 (5 unique balls)

jar 2 : B9, B7, B11, B15, B18 (5 unique balls)

jar 3 : B20, B1, B5, B17, B13 (5 unique balls)

jar 4 : B4, B5, B6, B3, B17 (5 unique balls)

jar 5 : B3, B10, B2, B14, B16 (5 unique balls)

and so on...< Snip>

It seems this gives you a total of 25>20 unique balls.

WWGD said:
It seems this gives you a total of 25>20 unique balls.
But only 20 unique ball types. The uniqueness of balls in a jar is trivial: a single ball cannot be in more than one jar, nor in the same jar more than once. There are 8 jars in total so a total of 40 (unique) balls.

Note that there are balls of ball type B1 for instance in both jar 1 and jar 3.

MrAnchovy said:
But only 20 unique ball types. The uniqueness of balls in a jar is trivial: a single ball cannot be in more than one jar, nor in the same jar more than once. There are 8 jars in total so a total of 40 (unique) balls.

Note that there are balls of ball type B1 for instance in both jar 1 and jar 3.

Ah, O.K, maybe it may be clearer to use the notation ## B_{1, i_1}, B_{2,i_2},..., B_{20, i_{20}} ## to denote the ball types, where ##i_j## ranges over the number of balls of type ##j##.

MrAnchovy said:
But only 20 unique ball types. The uniqueness of balls in a jar is trivial: a single ball cannot be in more than one jar, nor in the same jar more than once. There are 8 jars in total so a total of 40 (unique) balls.

Note that there are balls of ball type B1 for instance in both jar 1 and jar 3.
So you can't have the same ball type in the same jar

but you can have repeat of ball types for each jar

and there are 40 total different balls in 8 jars for some reason

As the first part of the problem has been shown to be trivial, and the OP has not returned to show any work on the second part it is probably not worth discussing any further.

1. What is the probability of picking a red ball from a jar with 10 red balls and 10 blue balls?

The probability of picking a red ball from this jar is 50%, since there are an equal number of red and blue balls.

2. If I pick a ball from the jar and then put it back, what is the probability of picking the same ball again?

The probability of picking the same ball again is the same as the first time, since each pick is an independent event and does not affect the contents of the jar.

3. How does the probability change if I pick 2 balls from the jar without replacement?

If 2 balls are picked without replacement, the probability of picking a red ball on the first pick is 10/20 or 50%. However, on the second pick, the probability will change depending on the outcome of the first pick. If a red ball was picked, the probability will now be 9/19 or approximately 47.4%.

4. What is the probability of picking all red balls from a jar with 10 red balls and 10 blue balls?

The probability of picking all red balls can be calculated by multiplying the probabilities of each pick. On the first pick, the probability is 10/20 or 50% for picking a red ball. On the second pick, the probability is now 9/19 or approximately 47.4%. This can be continued for all 10 picks, resulting in a probability of (10/20) * (9/19) * ... * (1/11) = approximately 0.0008 or 0.08%.

5. How can I calculate the expected number of red balls picked from a jar with 10 red balls and 10 blue balls?

The expected number of red balls can be calculated by multiplying the number of balls in the jar (20) by the probability of picking a red ball (10/20 or 50%). This results in an expected number of 10 red balls picked from the jar.

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