== THE SETUP ==
The table below shows the probability that Person N will draw a specific color ball out of a bag. Each person is given one opportunity to draw each color ball. For example, person 1 has one opportunity to draw a red ball (which he does with 25% probability of success), one opportunity to draw a green ball (which he does with 30% probability of success), etc. Each of these ball-drawing opportuity is independent.
Ball Color --> Red Green Yellow
Person 1 0.25 0.3 0.75
Person 2 0.35 0.65 0.4
== THE PROBLEM ==
I am trying to find the probability that, after all the ball drawing opportunities (there are six), the two people will share exactly zero (or one, or two, or three) balls of the same color.
== MY INITIAL SOLUTION ==
The probability they will share exactly zero is simple enough: (1-(.25 * .35)) * (1-(.3 * .65)) * (1-(.75 * .4)) = 0.514
The probability they will share all three is also straightforward: .25 * .35 * .3 * .65 * .75 * .4 = 0.0051
For the intermediate cases, the computation gets more difficult. The probability that the two people will share exactly one color of ball is 0.394, and can be computed as the sum of:
(A) The probability they both have red balls, but not green or yellow: (.25 * .35) * (1-(.3 * .65)) * (1-(.75*.4)) = 0.049
(B) The probability they both have green balls, but not red or yellow: (.3 * .65) * (1-(.25 * .35)) * (1-(.75 * .4)) = 0.124
(C) The probability they both have yellow balls, but not red or green: (.75 * .4) * (1-(.25 * .35)) * (1-(.3 * .65)) = 0.220
== MY REAL QUESTION ==
The problem is that this approach to finding the probability of two people sharing exactly one color of ball is that it requires summing comb(C,1) different probabilities, where C is the number of different colors of ball.
Is there a more direct way to compute the probability of sharing exactly one color of ball that could be used in a situation where there were, say, hundreds of different colors of balls.
The calculation you do for sharing none is not correct. Numerically, it gives 2.4175, which cannot be a probability. You should be multiplying the terms, at which point I get something close to $0.514$
Without symmetry between the chances of different colors of balls, I don't see a way easier than what you did. If you have hundreds of colors of balls, I would recommend a spreadsheet. With copy down you can handle hundreds or even a couple thousand easily.