question_answer

A speak truth in 30% of the case and B is 60% of the case. In what percentage of cases are likely to contradict each other in stating the same fact?

Answer:

Let E = Event of A speaking the truth and F = Event of B speaking the truth Then, \[P(E)=\frac{30}{100}=0.3\] and \[P(F)=\frac{60}{100}=0.6\] \[\Rightarrow \]   \[P(\bar{E})=1-P(E)=1-0.3=0.7\] and       \[P(\bar{F})=1-P(F)=1-0.6=0.4\] \[\therefore \] P(A and B contradict each other) = P(A speak truth and B speak false)    + P(A speak false and B speak truth) \[=P(E\cap \bar{F})+P(\bar{E}\cap F)=P(E)P(\bar{F})+P(\bar{E})P(F)\]                        [\[\because \] E and F are independent] \[=0.3\times 0.4+0.7\times 0.6\] \[=0.12\times 0.42=0.54\]