Question

A speaks the truth $3$ out of $4$ times, and $B5$ out of $6$ times : what is the probability that they will contradict each other in stating the same fact?

Answer 1

Probability that A speak truth,$P\left(A\right)=\frac{3}{4}$

Probability that B speak truth,$P\left(B\right)=\frac{5}{6}$

Probability that A and B lie is
$P\left(A^{{}^{\prime }}\right)$ lies$=\frac{1}{4}$ $P\left(B^{{}^{\prime }}\right)$ lies$=\frac{1}{6}$
Required probability$=P\left(A\right)\times P\left(B^{{}^{\prime }}\right)+P\left(A^{{}^{\prime }}\right)\times P\left(B\right)$
$=\frac{3}{4}\times \frac{1}{6}+\frac{1}{4}\times \frac{5}{6}$
$\Rightarrow \frac{8}{24}\Rightarrow \frac{1}{3}$