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Here, `S={1,2,3,4,5,6,7,8}" "therefore(S)=8` <br> (i) Let A be the event that the arrow rests at an odd number. <br> Then `A={1,3,5,7}" "therefore n(A)=4` <br> `(A)=(n(A))/(n(S))" "thereforeP(A)=4/8=1/2` <br> (ii) Let B the event that the arrow rests at a prime number. <br> Then `B={2,3,5,7}" " thereforen(B)=4` <br> `P(B)=(n(B))/(n(S))" "thereforeP(B)=4/8=1/2` <br> (iii) Let C be the event that the arrow rests at a multiple of 2. <br> Then `C={2,4,6,8}" "thereforen(C)=4` <br> `P(C)=(n(C))/(n(S))" "P(C)=4/8=1/2.` <br> In each of the cases the probability is `1/2.`