A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely incontradict each other while narrating some incident is
A speaks the truth 60% of the time, i.e., \( P(T_A) = 0.6 \).
B speaks the truth 50% of the time, i.e., \( P(T_B) = 0.5 \).
They will contradict each other when:
1. A speaks the truth and B lies.
2. A lies and B speaks the truth.
Calculating Probability of Contradiction:
Case 1: A speaks the truth (\( P(T_A) = 0.6 \)) and B lies (\( P(L_B) = 1 0.5 = 0.5 \)).
\[
P(T_A \cap L_B) = 0.6 \times 0.5 = 0.3
\]
Case 2: A lies (\( P(L_A) = 1 0.6 = 0.4 \)) and B speaks the truth (\( P(T_B) = 0.5 \)).
\[
P(L_A \cap T_B) = 0.4 \times 0.5 = 0.2
\]
Total probability of contradiction:
\[
P(\text{contradiction}) = P(T_A \cap L_B) + P(L_A \cap T_B) = 0.3 + 0.2 = 0.5
\]
Thus, the required percentage is \( 50\% \) or \( \frac{1}{2} \).
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