Total number of ways of throwing an unbiased die three times =6×6×6 ∵β1,β2,β3 satisfies ωβ1+ωβ2=−ωβ3 or ωβ1+ωβ2+ωβ3=ωβ3 or ωβ1+ωβ2+ωβ3=0 Then, β1,β2,β3 is a permutation of the numbers of the form 3k,3n+1,3m+2 from the set {1,2,3,4, 5,6} Thus, β1,β2,β3 can be chosen in 2×2×2×3! ways. Hence, required probability =