A bin contains 25 balls: 10 red, 8 yellow, and 7 blue. We draw three balls at random (without looking!) from the bin, and we will say that we "win" if our three balls represent exactly two colors. (That is, we "win" if we draw two balls of one color and another ball of a different color.)

What is the probability of winning in this particular game?

Since it was said "three bals at random (without looking), itīs intuitive that we are dealing with no replacement. The probability of winning is 1529/2300 or 66,478%.

We first need to count the number of possible outcomes there are to our game of drawing 3 balls from the bin.

Recall that the number of ways to choose k objects from a set of n objects is C(n,k) = n! / [k!(n-k)!].

In our case, we are choosing 3 balls from a set of 25 balls, so the number of choices is
C(25,3) = 25! / [3!22!] = [25*24*23] / [3*2] = 2300.

Now we just need to figure out how many of those possible choices will win the game. In order to win we need to do one of three things:

(1) draw exactly 2 red balls (with the third ball being some other color); (2) draw exactly 2 yellow balls; or (3) draw exactly 2 blue balls.

To choose exactly two red balls amounts to choosing two balls from the set of 10 red balls, followed by picking any of the 15 non-red balls as the third choice. So, the number of ways to choose exactly 2 red balls is
C(10,2)*15 = 45*15 = 675.

Similarly, the number of ways to choose exactly 2 yellow balls is
C(8,2)*17 = 28*17 = 476.

And the number of ways to choose exactly 2 blue balls is C(7,2)*18 = 21*18 = 378.

So, the total number of winning choices is
675 + 476 + 378 = 1529.

So, the probability of winning this game is 1529/2300 = 66,478%.

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