Correct Answer - A::B::C
Given, `P(A).P(B)=(1)/(6)(bar(A)).P(bar(B))=(1)/(3)`
`:." "[1-P(A)][1-P(B)]=(1)/(3)`
Let `" "P(A)=x" and "P(B)=y`
`implies" "(1-x)(1-y)=(1)/(3)" and "xy=(1)/(6)`
`implies" "1-x-y+xy=(1)/(3)" and "xy=(1)/(6)`
`implies" "x+y=(5)/(6)" and "xy=(1)/(6)`
`implies" "x((5)/(6)-x)=(1)/(6)`
`implies" "6x^(2)-5x+1=0`
`implies" "(3x-1)(2x-1)=0`
`implies" "x=(1)/(3)" and "(1)/(2)`
`:." "P(A)=(1)/(3)" or "(1)/(2)`