Let's assume a group of $80$ friends where $40$ of them plays baseball, $20$ of them plays football and $10$ of them plays both the game. Without any further information, if a random person is chosen, the probability of him playing football is $20/80 = 1/4$
Now let's say, we introduce one more information that the chosen person plays baseball. Now the sample space is halved (currently having only $40$ people) but so is the event space (currently $10$). So probability stays the same ($1/4$)
This example was shown to me by one of my friends and he claimed, here One person playing baseball and One person playing football is independent of each other. Because the occurrence of one event is not changing other's probability.
But I thought, if one event reduces other's possible outcomes (Here $10$ friends leaves the equation when I apply condition), how can they be considered independent.
If these events are indeed independent, the same example with different numbers, would surely change the probability. Will the same example then be considered having dependent events?
The Wikipedia article on Conditional Probability says,
If P(A|B) = P(A), then events A and B are said to be independent: in
such a case, knowledge about either event does not alter the
likelihood of each other
Then should I always calculate the answer of any probability question to determine what formula to use? Then how will I get to the answer in the first place?
Edit
Fine tuned terminologies to match with what I wanted to say
Best Answer
Yes, these events are independent. The definition of independent events only refers to the probabilities, not sample spaces: $P(A\cap B)=P(A)P(B)$.
You could argue that for any (nontrivial) events, conditioning on one event always changes the other's sample space. For example, if we roll two dice and define $A$ to be the event "the first die shows $6$" and $B$ to be the event "the second die shows $6$", then these are clearly independent. However, the real sample space for either event is all $36$ possible outcomes, and the event space for $B$ is $\{(1,6),(2,6),\ldots,(6,6)\}$. Once you condition on $A$, the sample space for $B$ changes to $\{(6,1),(6,2),\ldots,(6,6)\}$, and the event space changes to just $(6,6)$, but the probability hasn't changed and that is what makes them independent.
In a comment you suggested that the real sample space associated to event $B$ is just a set of size $6$ (the roll on the second die), and conditioning on $A$ doesn't change that. But you can do exactly the same thing with your initial example.
Divide the $80$ people into $20$ groups of $4$. Do this in such a way that the first $10$ groups all have four baseball players, one of whom also plays football, and the remaining $10$ groups have no baseball players but one football player each. The numbers you gave ensure this is always possible.
Now number the groups $1,\ldots,20$ and number the people in each group $1,2,3,4$, with the football player in each group being number $1$. You can choose a uniformly random person by choosing a number from $1,\ldots,20$ to pick a group, then independently choosing a number from $1,2,3,4$ to select a person from that group.
Now the event "plays baseball" is "the first number is at most $10$". The event "plays football" is "the second number is $1$". This is now exactly the same as the dice example - one event only refers to the first number and the other to the second.