SOLUTION: A box contains 4 red, 4 white and 5 green balls.Three balls are drawn from the box together. Find the probability that they may be i) all the same color ii) all different col

There are 4+4+5=13 balls.  So the number of ways to draw any 3 of them is
13C3 = (13∙12∙11)/(3∙2∙1) = 286 ways.

The number of ways they can be all red is 4C3 = (4∙3∙2)/(3∙2∙1) = 4 ways.

The number of ways they can be all white is 4C3 = (4∙3∙2)/(3∙2∙1) = 4 ways.

The number of ways they can be all green is 5C3 = (5∙4∙3)/(3∙2∙1) = 10 ways.

That's 4+4+10=18 ways they call all be the same color.

That's 18 ways out of 286 or a probability of 18/286 = 9/143

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Choose the red ball any of 4 ways.
Choose the white ball any of 4 ways.
Choose the green ball any of 5 ways.

That's 4∙4∙5 = 80 ways to choose the balls all different colors.

That's 80 ways out of 286 or a probability of 80/286 = 40/143

Edwin