A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident?

According to question, A speaks truth in 75% cases and B speaks truth in 80% cases.

Let, probability of A speaking truth be $$P(A)$$. So, $$P(A)=0.75$$ 

Also, probability of A speaking lie = $$P\left(A^c\right)$$ = $$1-P\left(A\right)$$ = $$1-0.75$$ = $$0.25$$

Probability of B speaking truth be P(B). So, $$P(B)=0.8$$

Also, probability of B speaking lie = $$P\left(B^c\right)$$ = $$1-P\left(B\right)$$ = $$1-0.8$$ = $$0.2$$

Now, A and B will contradict each other in two cases:

i.) A speaks truth and B speaks lie or,

ii.) B speaks truth and A speaks lie.

Now, probability that A speaks truth and B speaks lie = $$P\left(A∩B^c\right)=P\left(A\right)P\left(B^c\right)$$ = $$0.75\times\ 0.2=0.15$$

probability that B speaks truth and A speaks lie = $$P\left(B∩A^c\right)=P\left(B\right)P\left(A^c\right)$$ = $$0.8\times\ 0.25=0.20$$

So, probability that they will contradict each other = $$0.15+0.2=0.35$$