Suppose there are 3 red balls and 2 white balls in a bag. We want to pick out 2 balls without replacement. What's the probability of the 1st and 2nd balls are both red?
Solution 1: Use the conditional probability
Let $E_1=$ 1st ball red $E_2=$ 2nd ball red
then $P(E_1E_2)=P(E_2|E_1)\times P(E_1)=3/5 \times 2/4 = 3/10$
Solution 2: calculate the probability directly
Let $E=$ 1st and 2nd balls are red
There are $5!$ permutation of ball sequence. Of which $3\times 2 \times 3!$ are for $E$.
So $P(E)= 3 \times 2\times 3!/5!= 3/10$
The 2 results are identical just as expected. But why? Should we treat this coincidence as merely an evidence that the mathematical theory of probability is fortunately consistent with our practical experience? So that we can be more confident to apply the rule such as the conditional probability as long as no contradiction arise. Or is there any deep reason that ensures the results will inevitably be identical?
Here is a related question about the justification of Mathematical Theory of Probability.