Question

A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely to contradict each other while narrating some incident?

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{1}{3} \)

Hint:

To calculate the probability of contradiction, find the probabilities of each person speaking the truth or lying, and add the probabilities where they disagree.

Answer 1

The Correct Option is A

Step 1: Let the probability that A speaks the truth be \( P(A) = 0.6 \) and the probability that A lies be \( 1 - P(A) = 0.4 \). 
Similarly, let the probability that B speaks the truth be \( P(B) = 0.5 \) and the probability that B lies be \( 1 - P(B) = 0.5 \). 

Step 2: They will contradict each other in the following two cases: 

- Case 1: A speaks the truth, and B lies. This happens with probability \( P(A) \times (1 - P(B)) = 0.6 \times 0.5 = 0.3 \). 

- Case 2: A lies, and B speaks the truth. This happens with probability \( (1 - P(A)) \times P(B) = 0.4 \times 0.5 = 0.2 \). 

Step 3: The total probability of them contradicting each other is the sum of these two probabilities: 
\[ P(\text{contradiction}) = 0.3 + 0.2 = 0.5 \] Thus, the percentage of cases they are likely to contradict each other is \( 50% \), which corresponds to option (a) \( \frac{1}{2} \).