Calculating Probability of 2 Black Balls from 3-Ball Urn w/ Pascal Triangle

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firavia
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if we toss 4 coins and we want to know the probability of having 3 heads we can easily calculate that using pascal triangle .
I want to know if we have an Urn that contain 3 balls : 1 black , 1 red, 1 yellow.
and we want to draw 3 balls , but when we draw the first ball we put it back in the urn and thn we draw the 2nd ball which thn we put it back and finally we draw a third ball.

in this example we have 27 possible way of selecting 3 balls.

I want to know how can we calculate the probability of having at least 2 black ball , without using a tree diagram, probably wiht pascal triangle or any other formula with the explanation.

and you are very thankfull.

It is not a homework , I just want to know how this kind of problems works.
 
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You could set it out as a binomial tree if you fancied? Reducing the outcomes to Black and Not Black - i know you said you wanted to do it without a tree diagram but binomial trees never get old. (Much like probability questions involving urns)

This may, or may not, help you get a feel for it. If not try considering the "Events" that need to occur for you to get "at least 2 blacks" and then attach probabilities to these situations.
 
The probability of having at least 2 black balls means that you want to add the probability of getting 3 black balls (obviously 1 in 27) with the probability of having 2 black balls.

the other way to go is adding the probability of getting no black balls with the probability of getting 1 black ball, and subtracting that sum from 1.