Let [] be the sample space and [A ] be an event.

Question

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 :

Accepted Answer

$\Omega=$ 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, $P(A)=\frac{\text { Favourable volume }}{\text { Total volume }}$

Solutions · 1