# Doesn't order of drawing balls matter?

A bag contains an assortment of blue balls and red balls. If the balls are drawn at random, the probability of drawing two red balls is 5 times the probability of drawing two blue balls. Furthermore the probability of drawing one ball of each colour is six times the probability of drawing two blue balls. The number of red and blue balls in the bag is?

Let there be y red and x blue balls.

$$P(R) = \dfrac{^yC_2}{^{x+y}C_2}$$

$$P(B) = \dfrac{^xC_2}{^{x+y}C_2}$$

$$P(\text{one red and one blue}) = \dfrac{^2\times ^xC_1 \times ^yC_1}{^{x+y}C_2}$$

The above method (after making the necessary equations as per the given conditions) doesn't give the right answer.

But if I use:

$$P(\text{one red and one blue}) = \dfrac{ ^xC_1 \times ^yC_1}{^{x+y}C_2}$$ instead, I get the right answer.

I am unable to understand why the order of drawing balls isn't given any importance.

• I will point out that you are certainly allowed to make reference to the order in which they are drawn, but in that case both the numerator and the denominator should be counting ways in which the balls can be drawn where order matters. You would instead have $(x+y)(x+y-1)$ as the denominator rather than what you wrote in that case. – JMoravitz Nov 15 '18 at 19:21

When you use $$^{x+y}C_2$$ in the denominator you are counting the number of ways to draw two balls without respect to order, so the numerator should not reflect the order of drawing, either. You can check that without the factor $$2$$ you get three probabilities that add to $$1$$ as they should.