Question

Is it true that for any sets A and B, $P\left(A\right)\cup P\left(B\right)=P(A\cup B)$? Justify your answer.

Answer 1

No, it is not true.

Take A = {1, 2} and B = {2, 3}

Then $A\cup B$ = {1, 2, 3}

P(A) = {$\Phi ,$ {1}, {2}, {1, 2}}

P(B) = {$\Phi ,$ {2}, {3}, {2, 3}}

$\therefore P\left(A\right)\cup P\left(B\right)$ = {$\Phi ,$ {1}, {2}, {3}, {1, 2}, {2, 3}}...(i)

$A\cup B$ = {1, 2, 3}

$P(A\cup B)$ = {$\Phi$, {1}, {2}, {3}, {1, 2}, {2, 3}, {1,3}, {1, 2, 3}}...(ii)

From (i) and (ii), we have $P(A\cup B)\ne P\left(A\right)\cup P\left(B\right).$