Correct Answer – Option 1 : \(\rm \dfrac{xy}{xy+(1-x)(1-y)}\)

Concept:

Let A1, A2, …. , An be n mutually exclusive and exhaustive events of the sample space S and A is event which can occur with any of the events then

• \({\rm{P}}\left( {\frac{{{{\rm{A}}_{\rm{i}}}}}{{\rm{A}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}\)
•  

Calculations:

Consider, Let K be the event that both A and B  agree, 

T be the event that they both A and B speak the truth 

⇒ P(T) = xy

L be the event that they both A and B lie.
⇒ P(L) = (1 -x)(1 – y) 

To find :The probability that the statement is true = \(\rm P(\frac T L)\)

Let K be the event that both of them agree

 \(\rm P(\frac T L)\) = \(\rm \dfrac {P(T)P(\frac K T)}{P(T)P(\frac {K}{T})+P(L)P(\frac {K}{L}) }\)

⇒\(\rm P(\frac T L)\) = \(\rm \dfrac{xy}{xy+(1-x)(1-y)}\)