Question

For any sets $A$ and $B$, $P\left(A\right)$ is union of $P\left(B\right)=P\left(AunionB\right)$? (True/False) Justify your answer.

• A. True
• B. False

Answer 1

The correct option is B False

Verify the given statement

Let $A=\left\{0,1\right\}$ and $B=\left\{1,2\right\}$

$\therefore \left(A\cup B\right)=\left\{0,1,2\right\}$

Now,

$\begin{array}{rcl}P\left(A\right)& =& \left\{\varphi ,\left\{0\right\},\left\{1\right\},\left\{0,1\right\}\right\}\\ P\left(B\right)& =& \left\{\varphi ,\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\}\\ P\left(A\cup B\right)& =& \left\{\varphi ,\left\{0\right\},\left\{1\right\},\left\{2\right\},\left\{0,1\right\},\left\{0,2\right\},\left\{1,2\right\},\left\{0,1,2\right\}\right\}\\ P\left(A\right)\cup P\left(B\right)& =& \left\{\varphi ,\left\{0\right\},\left\{1\right\},\left\{2\right\},\left\{0,1\right\},\left\{1,2\right\}\right\}\\ \therefore P\left(A\right)\cup P\left(B\right)& \ne & P\left(A\cup B\right)\end{array}$

Hence, the statement is False.