Why doesn't the order matter in Probability

I got a horrible doubt in probability. Kindly help me. (in permutation and combination chapter)
Now our teacher has started probability Everything was almost clear unless and until I came with the second pic. When we are doing "10c1*15c1" the order is already being counted in this. Then why should I multiply it with two? Or in other language if we consider that order does not matter then it should be 10*15/2 and in denominator 25C2 And if we consider order matters then 10×15 (Here 1st one green 2nd one red /or / 1st one red and second one green has already been counted___as I did in the 1st pic) And in denominator 25×24

• Both approaches are correct. Either the order matters in both the numerator and denominator or the order does not matter in both the numerator and denominator. The result is the same. – Peter Sep 19 '18 at 9:35

Taking order into account

In this approach, it is the sequence of balls that are selected that matters.

The number of selections of two balls is $25 \cdot 24$ since we are selecting without replacement.

There are two favorable cases.

1. A red ball is selected, then a green ball is selected, which can occur in $10 \cdot 15$ ways.
2. A green ball is selected, then a red ball is selected, which can occur in $15 \cdot 10$ ways.

Therefore, the number of favorable cases is $10 \cot 15 + 15 \cdot 10$ and the probability of selecting one red ball and one green ball when two balls are selected is $$\Pr(\text{one red ball and one green ball}) = \frac{10 \cdot 15 + 15 \cdot 10}{25 \cdot 24}$$

Alternatively, we can use conditional probabilities. \begin{align*} \Pr(\text{one red ball and one green ball}) & = \Pr(R)\Pr(G \mid R) + \Pr(G)\Pr(R \mid G)\\ & = \frac{10}{25} \cdot \frac{15}{24} + \frac{15}{25} \cdot \frac{10}{24}\\ & = \frac{10 \cdot 15 + 15 \cdot 10}{25 \cdot 24} \end{align*} where

$\Pr(R) =$ the probability of selecting a red ball from the urn

$\Pr(G \mid R) =$ the probability of selecting a green ball from the urn given that a red ball has been selected from the urn

$\Pr(G) =$ the probability of selecting a green ball from the urn

$\Pr(R \mid G) =$ the probability of selecting a red ball from the urn given that a green ball has been selected from the urn

Not taking order into account

In this approach, it is which balls are selected that matters.

The number of ways two balls can be selected from $10$ red and $15$ green balls is $$\binom{25}{2}$$

There is one favorable case. We select a red ball and a green ball, which can occur in $$\binom{10}{1}\binom{15}{1}$$ ways since we must select one of the ten red balls and one of the ten green balls.

Thus, the probability of selecting one red ball and one green ball when two balls are selected is $$\Pr(\text{one red ball and one green ball}) = \frac{\dbinom{10}{1}\dbinom{15}{1}}{\dbinom{25}{2}}$$

The results are equal since taking order into account doubles both the numerator and denominator we obtain when we do not take order into account.

As long as you are consistent in taking order into account or not taking order into account in the numerator and denominator, you should get the same result for the probability.