Calculating Probability of Two Balls of Different Colors

[quasi426]

The problem first gives conditions: given that an urn contains n orange balls and n yellow balls, and two are selected at random.

The question was to determine the probability of getting two balls of different colors.

I did (1)*(n)/(2n-1)
Where 1 represented the probability of getting either color and the other term represented getting a ball of different color.

Anyway the answer was right but the book explained it in a way that i didn't understand. They said that there are n^2 chances of getting orange and then yellow and n^2 chances of getting yellow then orange. They divided this by the total possilbe which was 2n(2n-1)

[n^2+n^2]/2n(2n-1)

Can someone explain to me why n^2 equals what it equals. Thank you.

 

[EnumaElish]

There are n orange balls and anyone of them can be selected. So one can select orange_1, orange_2, ..., orange_n. Same for yellow: yellow_1, ..., yellow_n. Think of a table with n rows and n columns. Each row represents an orange ball. Each column represents a yellow ball. The number of different (orange, yellow) combos is n*n = n^2.

 

[vaishakh]

You get 2n^2 in the numerator which is the nth multiple of 2n in the denominator. why didn't you simplify it. you can't simplify a^2 + b^2. but why not when both the variables are same. thus the answer is the same
how did you get the answer? the actual method is the one given by the book. you say as if the book has done it in a differant. i don't know what is the method you stick. probably i would be able to point out again the similarity between two.