Correct Answer - C
Let `x inP(AnnB)`
`iffxsube(AnnB)`
`iffxsubeAandxsubeB`
`iffx inP(A)andx inP(B)`
`iffx inP(A)nnP(B)`
`therefore P(AnnB)subeP(A)nnP(B)`
and `P(A)nnP(B)subeP(AnnB)`
Hence, `P(A)nnP(B)=P(AnnB)`
Now, consider sets A = {1}, B = {2} implies `A uu B` = {1, 2}
`therefore P(A) = {phi, {1}}, P(B)={phi, {2}}`
and `P(AuuB)={phi {1}, {2}, {1,2} ne P(A)uuP(B)}`
Hence, Statement-1 is true and Statement-2 is false.