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There are dependent and independent events.
In the first variant, cards are taken from the deck without being returned.
These are dependent events.
In the second option:
The card was taken out, returned to the deck, shuffled, taken out again, and so on.
These are independent events.
My answer, by the way, has to do with the condition of dependent events.
рAABB = 2/30 * 1/29 * 2/28 * 1/27 = 1/164430
pBBAA = 2/30 * 1/29 * 2/28 * 1/27 = 1/164430
pABBA = 2/30 * 2/29 * 1/28 * 1/27 = 1/164430
pBAAB = 2/30 * 2/29 * 1/28 * 1/27 = 1/164430
pABAB = 2/30 * 2/29 * 1/28 * 1/27 = 1/164430
pBABA = 2/30 * 2/29 * 1/28 * 1/27 = 1/164430
for the rest it's the same
To satisfy the condition of point 2, any sequence will do:
or AABB, or BBAA, or AABB, or BBAA, and so on
therefore, all the probabilities found are summed up
P = 1/ 164430 * 36 ≈ 0.0002189381
The second way to find this probability is to use the factorial.
And this strict sequence will no longer be important.
See the topic "Discrete Distributions"
Pay attention to the denominators.
Here I have a decision when the cards are eliminated from play.
Moreover, it can be solved in two ways.
For independent events there will be a different solution.
Everything is fine as it is, and it will get even better.
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