Question about determining probability

[Mr.V.]

Hi!
I have a data set of ~5000 unique elements.
From that set I have 2 subsets that are not mutually exclusive. For example if the elements are letters from A-Z, the first set could be A, B, C, D, E, F, and G the second set could be E, F, G, H, I, and J.
Here's the question...
The first subset has 100 elements randomly chosen from the 5000. The second subset has 100 elements randomly chosen from the 5000. Of interest is that 10 of the elements from subset1 are also in subset 2.
What is the probability of that happening?
Here's my logic so far, though I'm not sure I'm right.
If we use subset 1 as the reference:
If subset 1 had 1 element, the probability of getting any 1 element in subset 2 is:
$$\frac{100}{5000}$$ or $$\frac{1}{50}$$. The probability of getting 2 elements in a set of 2 is: $$\frac{100}{5000} * \frac{99}{4999}$$ ... the probability of getting 10 if subset 1 were only 10 elements would be $$\frac{100!}{90!}*\frac{4990!}{5000!}$$ which is roughly $$(\frac{1}{50})^{10} = 1.024*10^{-17}$$
However since I have a set of 100, I need to include the chance of getting that set of 10 in many different ways...I think I should use combinations correct? So since I had a set of 100, and I want a subset of 10, there are $$\frac{100!}{10!*90!}$$ different ways of ordering that set...
So is the correct answer...
$$(\frac{1}{50})^{10} * \frac{100!}{10!*90!} = 0.00018$$
?
Thanks for your help!

 

[NateTG]

Hmm.

Lets say we have 1000 objects, and 20 are in a particular subset - call it set A, and we want to know what the odds are of picking a subset B of 20 elements at random so that exactly 2 of these elements is in A.

Clearly, there are 1000 choose 20 ways to choose B.
Now, if B is to have two elements from A, then there are 20 choose 2 ways to pick those two elements, and, since B will also have 18 elements from the remaining 1980 objects, there are 1980 chose 18 ways to pick those.

Thus, the probablity of getting exactly 2 elements would be:
$$\frac{20 \rm{C}2 \times 1980 \rm{C}18}{2000 \rm{C} 20}$$

 

[Architeuthis Dux]

I would approach this question by first clarifying the goal of determining the probability. Is the goal to find the probability of getting exactly 10 elements that are also in subset 2, or is it to find the probability of getting at least 10 elements that are also in subset 2? This distinction is important because the probability will be different for each scenario.

Assuming the goal is to find the probability of getting exactly 10 elements that are also in subset 2, your logic and calculations seem to be correct. You have correctly used the concept of combinations to account for the different ways in which the 10 elements can be chosen from the set of 100 elements in subset 1.

However, if the goal is to find the probability of getting at least 10 elements that are also in subset 2, then the calculation would be slightly different. In this case, you would need to calculate the probability of getting 10, 11, 12,...,100 elements that are also in subset 2, and then add all these probabilities together to get the overall probability. This would require more calculations, but the concept of combinations can still be applied to each scenario.

In conclusion, your approach and calculations seem to be correct, but it is important to clarify the goal of determining the probability before proceeding with the calculations.