x speaks truth in 60% and y in 50% of the cases. The probability that…

x → event that x speaks truth

y → event that y speaks truth

$\bar{x}$ → x doesn't speak truth

$\bar{y}$ → y doesn't speak truth

$P(\bar{x}) =1-P(x)$

$=\frac{1-\frac{6}{10}}{10}=\frac{4}{10}$

$P(\bar{y}) =1-P(y)$

$=1-\frac{1}{2}=\frac{1}{2}$

so P(x) = 60% = $\frac{60}{100} = \frac{6}{10}$

so P(y) = 50% = $\frac{50}{100} = \frac{1}{2}$

so both contradict each other when 

x → speaks truth , y → speaks truth

→ $P = P(x) P(\bar{y})+P(\bar{x})P(y)$

$P = \frac{6}{10}×\frac{1}{2} + \frac{4}{10}×\frac{1}{2} = \left(\frac{6+4}{10}\right) \times \frac{1}{2}=\frac{1}{2}$