Conditional Probability - How to calculate of an event occurring based on previous event data?

[luthius]

I believe this is a conditional or a priori probability problem.
There are 15 balls of different colors in a basket.
In each choice/attempt, WITH REPLACEMENT, 05 balls are chosen randomly.

1- What is the probability of choosing 03 balls from the previous attempt in the second attempt? I mean, repeat 03 colors that were chosen in the previous attempt (First choice/Draw/attempt)?
2- Considering the next attempt, what is the probability of choosing again 3 colors from the previous attempt (Second choice/Draw/attempt)?

I would like the result to be presented with the steps preferably with the construction of functions to reach the result.
Thank you very much in advance guys.

 

[Dr.Peterson]

I would like the result to be presented with the steps preferably with the construction of functions to reach the result.

That's not quite how we work:

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We'd like to see your thoughts first, so we can use them as the basis for your ultimate solution, rather than just give you ours.

In each choice/attempt, WITH REPLACEMENT, 05 balls are chosen randomly.

For clarity, are you saying that the 5 balls in each round are chosen with replacement, or that within a round, they are chosen without replacement (so that they are distinct balls), and then they are all put back for the next round?

1- What is the probability of choosing 03 balls from the previous attempt in the second attempt? I mean, repeat 03 colors that were chosen in the previous attempt (First choice/Draw/attempt)?
2- Considering the next attempt, what is the probability of choosing again 3 colors from the previous attempt (Second choice/Draw/attempt)?

How are these different? Assuming you're asking about the third round, as I'm calling it, isn't its relationship to the second the same as that of the second to the first?

If your central question is, is question 2 asking for a conditional probability, that's a matter of the language used, and the language is not very clear. You need to ask whoever posed the questions.

 

[luthius]

A: 5 balls in each round are chosen with replacement, meaning that for the second round they are all put back again inside the basket.

By saying that, 05 distinct balls are chosen out of 15. In each round we have as combinations of colous an amount of 3003 different ways.
[math]C \binom{15}{5}=3003[/math]
Considering the colours to be chosen are defined before someone choose them, the probability is:
1/3003
My thoughts are:

In case to at least 3 balls from the previous attempt be chosen again, how would it be? Something like that?
[math]1/{C \binom{15}{5}*C \binom{15}{2}}[/math] ?????

I cannot figure it out.

 

[Dr.Peterson]

A: 5 balls in each round are chosen with replacement, meaning that for the second round they are all put back again inside the basket.

This is still ambiguous, but I think you mean this:

In each round, 5 balls are chosen without replacement, and then they are all put back again for the second round (so that replacement occurs between rounds, not within a round).​

Am I right? If not, then you are saying that after each individual ball is picked, its color is noted and it is immediately put back (which is what "chosen with replacement" means).

[math]C \binom{15}{5}=3003[/math]

You are using a mixed notation; we don't use the letter C with the form you used. The following all mean the same thing: [imath]_{15}C_5[/imath], [imath]C(15,5)[/imath], and [imath]\binom{15}{5}[/imath] (and a few others!).

Considering the colours to be chosen are defined before someone choose them, the probability is:
1/3003
My thoughts are:

In case to at least 3 balls from the previous attempt be chosen again, how would it be? Something like that?
[math]1/{C \binom{15}{5}*C \binom{15}{2}}[/math] ?????

I cannot figure it out.

You are right that there are 3003 equally likely outcomes from the first round; but your first answer of 1/3003 would be the probability that all 5 would match.

In your second attempt, you are right to use the 3003 as the denominator; the numerator should be the number of ways to choose 3 balls from the first set of 5, and 2 others. Your numerator (second factor) is just the number of ways to choose 2 of any color.

(I'm assuming that "3 colors that were chosen in the previous attempt" means exactly 3, and not "at least 3".)

So, how many ways are there to choose 3 from the first 5 and 2 from the other 10?

 

[luthius]

So, how many ways are there to choose 3 from the first 5 and 2 from the other 10?

Do you mean [math]\binom{5}{3} * \binom{10}{2} = 450[/math]By saying it can we consider a general approach for different quantities of balls from the previous one represented by x variable below?
[math]P=1/\binom{N}{k}[/math]
[math]P(x)=(\binom{k}{x} * \binom{N-k}{k-x})/\binom{N}{k}[/math]
Am I correct in my thoughts?

If its correct, developing a table of possibilities of choosing X balls from previous attempts we would have something like this:



I'm still confused because if I sum all the probabilities its NOT 100%. It is resulting in 91.6084...%

 

[Dr.Peterson]

Do you mean [math]\binom{5}{3} * \binom{10}{2} = 450[/math]By saying it can we consider a general approach for different quantities of balls from the previous one represented by x variable below?
[math]P=1/\binom{N}{k}[/math]
[math]P(x)=(\binom{k}{x} * \binom{N-k}{k-x})/\binom{N}{k}[/math]
Am I correct in my thoughts?

Yes that's all correct, assuming N=number of balls, k=number selected on each round, x=number matching previous round.

If its correct, developing a table of possibilities of choosing X balls from previous attempts we would have something like this:

View attachment 29240

I'm still confused because if I sum all the probabilities its NOT 100%. It is resulting in 91.6084...%

You are wise to check this way. But you omitted one case: x=0. Include that, and the sum will be 100%!

Good work.

 

[luthius]

Yes that's all correct, assuming N=number of balls, k=number selected on each round, x=number matching previous round.


You are wise to check this way. But you omitted one case: x=0. Include that, and the sum will be 100%!

Good work.

Thank you very very much... Problem solved.