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 :

Question

Accepted Answer

Answer: B. both (S1) and (S2) are false \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}ight\} = Selecting a point on the wire which is at \left\{\frac{1} {2}ight\} or 0.5 As wire is an 1-D object so from geometrical probability P(A)=\frac{	ext { Favourable Length }}{	ext { Total Length }} Here total length of wire = 1 unit and point has zero length so point A at \left\{\frac{1} {2}ight\} or 0.5 has length = 0. 	herefore Favorable length = 0 	herefore \mathrm{P}(\mathrm{A})=0 {	ext { but }} \mathrm{A} 
eq \phi Now \overline{\mathrm{A}} = [0,1] - \left\{\frac{1} {2}ight\} So, length of \overline{\mathrm{A}} = Length of entire wire - Length of point A = 1 	herefore \mathrm{P}(\overline{\mathrm{A}})=1 {	ext { but }} \overline{\mathrm{A}} 
eq \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{	ext { Favourable Length }}{	ext { Total Length }} 2. For 2-D object, P(A)=\frac{	ext { Favourable Area }}{	ext { Total Area }} 3. For 2-D object, P(A)=\frac{	ext { Favourable volume }}{	ext { Total volume }}

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 :

Answer

Explanation

More JEE Math questions