Probability of drawing three different colors

Question

You have a closed box with 10 red balls, 10 white balls and 10 blue balls. (Otherwise the balls are identical.) Without looking in the box, you draw out exactly three balls.

 

What is the probability that the three balls will all be a different color (i.e. 1 red, 1 white and 1 blue)? The order of color does not matter, only that all three balls are different.

Answer 1

P(3 diff colors) = [ 10C1•10C1•10C1] ÷ 30C3</div the notation nCr gives # of ways of getting a choice of r things, from possible n. the first factor 10C1 tells you that there's one ten possible choices for the first color desired; the other factors are for the other two colors; the value of this numerator is 10 cubed. The denominator tells you how many ways to choose a combination of three balls, from 30 available. This (denominator) value equals 30! divided by [27! 3!] which equals 30(29)(28)/6. Thus you can calculate the answer as a rational (fraction) or decimally).