Re: A problem for complete beginners

I proposed this problem in order to make clear what is a sample with replacement and without replacement. These are simple concepts but even so confusing for some people.
A better definition(I think):
_____If the items drawn yet are not able to be again we say that is without replacement. If prior to each drawn they have the same probability to be chosen we say the drawings are with replacement.________
(see below an improved definition).
The Ann’s problem just intend to show that * the simple fact that the drawn cards be closed to the remaining does not allow us to consider we have a with replacement case * . On the contrary, they are without replacement.
WHY?
Because we stated the sampled cards are only the three on the top. All yet sampled are discriminated, cannot be drawn again.
The probability we are

____P=3*p=3*(26/52)*(25/51)*(26/50)

Where p is the case BBR (black, black, red).
__prior the 1st drawn they are 26 black out of 52
__prior the 2nd they are 25 out of 51
__prior the last dawn they are 26 red out of 50.

However the cases RBB and BRB must be considered, therefore 3 chosen 2 multiply p.

A remark:

Some elementary books say (Hog and Tanis, Probability and Statistical Inference), I quote:

* Sampling with replacement (definition): occurs when an object is selected and then replaced before the next object is selected *.

This is not sufficient because the replaced item must have the same probability before and after replacement (or a known one in the recapture of an animal in nature), BUT COULD BE different from the others.

Tentatively improving:

_p(after) = 0 : drawing without replacement
_p(before) not equal to 0: drawing with replacement But it could be ___p(after) not equal to p(before).

I apologize if the problem is worthless.


____________licas_@xxxxxxxxxxx
.

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