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Total balls =4+6+5=15= 4+6+5=15. Total unordered pairs =(152)=105=\binom{15}{2}=105. Pairs of the same color:

Red–Red: (42)=6\binom{4}{2}=6

Blue–Blue: (62)=15\binom{6}{2}=15

Green–Green: (52)=10\binom{5}{2}=10

Sum =6+15+10=31= 6+15+10=31.

Therefore, pairs of different colors =105−31=74= 105 - 31 = 74. Probability =74/105= 74/105. This fraction is already in lowest terms (74 = 2×372 \times 37; 105 = 3×5×73 \times 5 \times 7). An equivalent route uses the complement with sequential probabilities: P(different) =1−[415⋅314+615⋅514+515⋅414]=1−31105=74105= 1 - \big[\tfrac{4}{15}\cdot\tfrac{3}{14} + \tfrac{6}{15}\cdot\tfrac{5}{14} + \tfrac{5}{15}\cdot\tfrac{4}{14}\big] = 1 - \tfrac{31}{105} = \tfrac{74}{105}. Common pitfalls include forgetting “without replacement,” which keeps denominators at 15 then 14, or double-counting ordered pairs; using combinations avoids that trap. Sanity check: different-color pairs should be much more likely than same-color pairs since there are three colors with uneven counts; 74105≈0.704\tfrac{74}{105}\approx 0.704 is reasonable.