Two reporters A and B narrate the same incident. A tells the truth in 60% of cases; B in 80% of cases. What is the probability they contradict each other (one says true while the other says false)?

Introduction / Context:
They contradict when exactly one speaks truth and the other lies about the same fact. With independent truth-telling behavior, compute the two mutually exclusive ways this happens and add them.

Given Data / Assumptions:

• P(A truthful) = 0.6 → P(A lies) = 0.4.
• P(B truthful) = 0.8 → P(B lies) = 0.2.
• Independence between A and B.

Concept / Approach:
P(contradict) = P(A true & B false) + P(A false & B true).

Step-by-Step Solution:
P(A true & B false) = 0.6 × 0.2 = 0.12.P(A false & B true) = 0.4 × 0.8 = 0.32.Total = 0.12 + 0.32 = 0.44 = 44%.

Verification / Alternative check:
P(agree) = P(both true) + P(both false) = (0.6×0.8) + (0.4×0.2) = 0.48 + 0.08 = 0.56, and 1 − 0.56 = 0.44 (contradict), consistent.

Why Other Options Are Wrong:
70% is unrelated; 36% or 22% come from partial products; 40% cannot result from these parameters.

Common Pitfalls:
Multiplying 0.6 and 0.8 or adding them directly—neither operation models contradiction correctly.

Final Answer:
44%