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B
P(A)∩{A}=ϕ
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C
P(A)∩{A}=P(A)
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D
P(A)∪{A}=P(A)
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Solution
The correct options are BP(A)∩{A}={A} DP(A)∪{A}=P(A) Consider A={a,b} then P(A)={ϕ,{a},{b},{a,b}} ∴P(A)⋂{A}={ϕ,{a},{b},{a,b}}⋂{{a,b}}={{a,b}}={A} P(A)⋃{A}={ϕ,{a},{b},{a,b}}