QJEEM23Ma3a25: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3a25 Solution: Let Ω be the sample space and A ⊆ Ω be an event. Given below…

$\Omega=$ sample space

$\mathrm{A}=$ be an event

$\Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$

$\mathrm{A}=\left\{\frac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\frac{1} {2}\right\}$ or 0.5

As wire is an 1-D object so from geometrical probability

$P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$

Here total length of wire = 1 unit

and point has zero length so point A at $\left\{\frac{1} {2}\right\}$ or 0.5 has length = 0.

$$\therefore$$ Favorable length = 0

$$\therefore$$ $\mathrm{P}(\mathrm{A})=0 {\text { but }} \mathrm{A} \neq \phi$

Now $\overline{\mathrm{A}}$ = $[0,1]$ - $\left\{\frac{1} {2}\right\}$

So, length of $\overline{\mathrm{A}}$ = Length of entire wire - Length of point A = 1

$$\therefore$$ $\mathrm{P}(\overline{\mathrm{A}})=1 {\text { but }} \overline{\mathrm{A}} \neq \Omega$ .

Then both statements are false.

Attention : According to NTA option A is correct. Which is wrong. That is proven here using geometrical probability.

Note :

Geometrical probability :

1. For 1-D object, $P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$

2. For 2-D object, $P(A)=\frac{\text { Favourable Area }}{\text { Total Area }}$

3. For 2-D object, $P(A)=\frac{\text { Favourable volume }}{\text { Total volume }}$