A speaks truth in 60% and Bin 80% of the cases. In what percentage of cases are they likely to contradict each other narrating the same incident?

Options

A. `44%`
B. `36%`
C. `64%`
D. `48%`

Answer

To solve the problem, we need to calculate the probability that A and B contradict each other when narrating the same incident. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the probabilities of telling the truth and lying
- A speaks the truth 60% of the time, so the probability of A telling the truth (P(A)) is:
\[ P(A) = \frac{60}{100} = 0.6 \]
- Therefore, the probability of A lying (P(A')) is:
\[ P(A') = 1 - P(A) = 1 - 0.6 = 0.4 \]

- B speaks the truth 80% of the time, so the probability of B telling the truth (P(B)) is:
\[ P(B) = \frac{80}{100} = 0.8 \]
- Therefore, the probability of B lying (P(B')) is:
\[ P(B') = 1 - P(B) = 1 - 0.8 = 0.2 \]

### Step 2: Calculate the probabilities of them contradicting each other
There are two scenarios where A and B contradict each other:

1. **Case 1**: A tells the truth and B lies.
- The probability of this happening is:
\[ P(A) \times P(B') = 0.6 \times 0.2 = 0.12 \]

2. **Case 2**: A lies and B tells the truth.
- The probability of this happening is:
\[ P(A') \times P(B) = 0.4 \times 0.8 = 0.32 \]

### Step 3: Add the probabilities of the two cases
To find the total probability that A and B contradict each other, we add the probabilities from both cases:
\[ P(\text{contradiction}) = P(A) \times P(B') + P(A') \times P(B) = 0.12 + 0.32 = 0.44 \]

### Step 4: Convert the probability to a percentage
To express this probability as a percentage, we multiply by 100:
\[ \text{Percentage of contradiction} = 0.44 \times 100 = 44\% \]

### Final Answer
Thus, A and B are likely to contradict each other in **44%** of the cases.

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