Why the two probability results are the same?

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

$P(E_1)=3/5$ $P(E_2|E_1)=2/4$

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.

• $\frac{3 \cdot 2 \cdot 3!}{5!} = \frac{3 \cdot 2 \cdot (3 \cdot 2 \cdot 1)}{5 \cdot 4 \cdot 3 \cdot2 \cdot1}= \frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ So both formulas are the same which you would expect as they are both trying to solve the same problem. – Warren Hill Jan 29 '14 at 9:03
• @Warren Hill Thanks but the reasoning is not convincing enough. – smwikipedia Jan 29 '14 at 9:05

A "deep reason" is probably how the conditional probability formula $$P(E_2|E_1) = \frac{ P(E_1 \cap E_2)}{P(E_1)},$$ which you used in Method 1, arrives when using the second method.
Using the second approach, the LHS is $$P(E_2|E_1) = \frac{\text{number of ways to arrange the remaining 4 balls such that the first of them is red, given that the first ball was red}}{\text{number of ways to arrange the remaining 4 balls given that the first ball was red}}$$ but hey, this becomes $$P(E_2|E_1) = \frac{\text{number of ways to arrange the 5 balls such that the first two are red}}{\text{number of ways to arrange the 5 balls such that the first ball is red}},$$ dividing both the numerator and denominator by "number of ways to arrange the 5 balls", we get $P(E_1 \cap E_2)$ and $P(E_1)$, respectively. Thus, $$P(E_2|E_1) = \frac{ P(E_1 \cap E_2)}{P(E_1)}$$
• I wouldn't say it is luck. The space $X$ of all elementary events (arrangements of 5 balls) is divided into four parts, $E_1 \cap E_2$, $E_1 \cap E_2^C$, $E_1^C \cap E_2$, and $E_1^C \cap E_2^C$. If every elementary event has equal probability, we can use division to get the probabilities. For conditional probability $|E_1$, we can think that we are living in a universe where $X$ contains only the elementary events for which $E_1$ is true, i.e. $X$ is divided into $E_1 \cap E_2$ and $E_1 \cap E_2^C$, and can use division to get the probability of $E_2$ in that universe. – JiK Jan 29 '14 at 9:31