A speaks the truth in 70% cases and B speaks the truth in 84% cases. What is the probability that they contradict each other on a statement?

Correct option is D

​$$\begin{aligned}&\text{We are given:} \\[5pt]&\bullet\ \text{A speaks the truth with probability } P(A_T) = 0.70,\ \text{so lies with probability } P(A_L) = 1 - 0.70 = 0.30 \\[5pt]&\bullet\ \text{B speaks the truth with probability } P(B_T) = 0.84,\ \text{so lies with probability } P(B_L) = 1 - 0.84 = 0.16 \\[15pt]&\textbf{We are to find the probability that A and B contradict each other, i.e., one tells the truth and the other lies.} \\[8pt]&\text{This happens in two mutually exclusive cases:} \\[8pt]&1.\ \text{A tells the truth and B lies:} \\&P(A_T \cap B_L) = P(A_T) \cdot P(B_L) = 0.70 \cdot 0.16 = 0.112 \\[10pt]&2.\ \text{A lies and B tells the truth:} \\&P(A_L \cap B_T) = P(A_L) \cdot P(B_T) = 0.30 \cdot 0.84 = 0.252 \\[15pt]&\text{Total probability that A and B contradict each other:} \\&P(\text{contradiction}) = 0.112 + 0.252 = {0.364}\end{aligned}$$​​