Otdfbt

The Problem:

An urn contains 3 red and 4 black balls and another contains 4 red and 5 black.

A random ball is chosen from the first urn and is inserted into the second urn.

After this a random ball is chosen from the second urn.

Consider the events:
$A$ : "first ball is red", $B$ : "second ball is red".

Find the Probability $P_A(B)$ where "$P_A(B)$" means the probability that $B$ happens if $A$ has happened (i.e. the conditional probability of $B$ given $A$).

This is a simple problem right?

Why I'm confused:

But I'm confused. We work in the probability space $(Omega,Sigma,P)$
(in this case $Sigma=mathcal P(Omega)$), and $P$ is a probability that means is a function which respect kolmogorov axioms.

$P_A(B)=1/2$ (because if $A$ happened then in the second urn we will have 5 red balls and 5 black balls)

My question:

Here is my question: If $P_A(B)=fracP(A cap B)P(A)$ by definition (and $P $ is a function, we don't know what function just this function respect Kolmogorov axioms). How we came to the conclusion $P_A(B)=1/2$? Using
$$fractextNumber of Favorable OutcomestextTotal Number of Possible Outcomes?$$
This doesn't make sense for me. $P$ is fixed from the beginning, then we must find $P_A(B)$ using definition to be rigorous. I need a rigourous proof.

The issue that you raised is a legitimate one. The calculation leading to $frac12$ in your question (and the calculations in both of the answers available as I write this) use the assumption that, when a ball is drawn from an urn, each of the balls in that urn is equally likely to be drawn. That assumption is part of what people usually understand by the phrase "a random ball is chosen", which occurs twice in the problem. Sometimes people express this more explicitly by saying that a ball is chosen "uniformly at random".

It is certainly possible to have physical situations where this assumption is incorrect. For example, if all the red balls are much smaller than the blue ones, settle to the bottom of the urn, and are almost impossible to find and grab. Then pulling a "random" ball from such an urn would not make all the balls equally probable.

The general theory of probability, with its arbitrary probability function $P$, is designed to apply to all situations, even ones where the sizes of the balls or some other irregularities make some balls more likely to be chosen than others. To solve problems like the one you quoted, one needs additional information about the function $P$. In the case at hand, "a random ball is chosen" was intended to supply the additional information that all balls in an urn are equally likely to be chosen.

(And $P$ is a function, we don't know what function just this function respect Kolmogorov axioms)

We do know more.   We know that our $mathsf P$ is the probability measure for the events of the described process.   It has to obey the Kolmogrov axioms, but also be evaluated according to reasoning about the selection proceedure.

Here too our $Omega$ is not just an abstraction, but a set of outcomes for this ball selection process. Likewise our $mathcal F$ is the event algebra for this particular probability space, and our events, $A$ and $B$, are contained with in it.

$mathsf P_A(B)$, which is more commonly written $mathsf P(Bmid A)$, is the probability for selecting one from the red balls in the second urn when given that one red ball was added to the four red, and five black, balls originally in that urn.   This is just the probability for obtaining one from five red balls among ten balls. Hence $mathsf P_A(B)$ equals $tfrac 510$.

$mathsf P(Acap B)$ is the probability for selecting one from the three red balls among the seven balls in the first urn to place in the second urn, then selecting one from the four-plus-one red balls among the nine-plus-one balls in the second urn.   So $mathsf P(Acap B)$ equals $tfrac 37cdottfrac 510$, which is $tfrac 1570$.

$mathsf P(A)$ is the probability of selecting one from the three red balls among the seven balls in the first urn, which is just $tfrac 37$.   So clearly $mathsf P(Acap B)/mathsf P(A)$ does equal $tfrac 510$ as anticipated.

$P(A)=frac37$. What is $P(Acap B)$? First we draw a red ball from urn $1$ with probability $frac37$. Then we independently draw a red ball from urn $2$ with probability $5over10$. So $$P(Acap B)=frac37cdot5over10=3over14$$ and $$P_A(B)=3/7over3/14=frac12.$$

Initially, we have $70$ outcomes: $7$ ways to draw a ball from urn $1$ and $10$ ways to draw a ball from urn $2$. Once we say that a red ball is drawn from urn $1$ we discard $40$ of those outcomes, and there are only $30$ left. I think you may be a bit confused when you say, "$P$ is fixed from the beginning." This is true of course, but $P_A$ is not the sam as $P$. Once we condition on $A$ we have a different probability space. The events are different, since those where a black ball is drawn first are gone. Of the $30$ remaining events $15$ are favorable: draw one of the $3$ red balls from urn $1$ and then draw one of the $5$ red balls from urn $2$.