Let $\Omega$ be the sample space and $\mathrm{A \subseteq \Omega}$ be an event.

Given below are two statements :

(S1) : If P(A) = 0, then A = $\phi$

(S2) : If P(A) = 1, then A = $\Omega$

Then :

$\Omega=$ sample space<br/><br/> $\mathrm{A}=$ be an event<br/><br/> $ \Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$<br/><br/> $\mathrm{A}=\left\{\frac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\frac{1} {2}\right\}$ or 0.5 <br/><br/>As wire is an 1-D object so from geometrical probability <br/><br/>$P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$ <br/><br/>Here total length of wire = 1 unit <br/><br/>and point has zero length so point A at $\left\{\frac{1} {2}\right\}$ or 0.5 has length = 0. <br/><br/>$\therefore$ Favorable length = 0 <br/><br/> $\therefore$ $\mathrm{P}(\mathrm{A})=0 {\text { but }} \mathrm{A} \neq \phi$ <br/><br/>Now $\overline{\mathrm{A}}$ = $[0,1]$ - $\left\{\frac{1} {2}\right\}$ <br/><br/>So, length of $\overline{\mathrm{A}}$ = Length of entire wire - Length of point A = 1 <br/><br/> $\therefore$ $\mathrm{P}(\overline{\mathrm{A}})=1 {\text { but }} \overline{\mathrm{A}} \neq \Omega$.<br/><br/> Then both statements are false. <br/><br/><b>Attention :</b> According to NTA option A is correct. Which is wrong. That is proven here using geometrical probability. <br/><br/><b>Note :</b> <br/><br/>Geometrical probability : <br/><br/>1. For 1-D object, $P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$ <br/><br/>2. For 2-D object, $P(A)=\frac{\text { Favourable Area }}{\text { Total Area }}$ <br/><br/>3. For 2-D object, $P(A)=\frac{\text { Favourable volume }}{\text { Total volume }}$