Sample space A dice is thrown thrice, n(s)=6×6×6.Favorable events ωr1​+ωr2​+ωr3​=0i.e. (r1​,r2​,r3​) are ordered 3 triples which can take values,and each can be arranged in 3 ! ways =6∴n(E)=8×6⇒P(E)=6×6×68×6​=92​