X speaks truth in 60% and Y in 50% of the cases. The probability that they contradict each other narrating the same incident is

Options

A. `1/4`
B. `1/3`
C. `1/2`
D. `2/3`

Answer

To solve the problem, we need to find the probability that X and Y contradict each other when narrating the same incident.

### Step-by-step Solution:

1. **Understand the probabilities of speaking the truth:**
- X speaks the truth in 60% of the cases, which can be expressed as:
\[ P(\text{X speaks truth}) = 0.6 = \frac{6}{10} \]
- Therefore, the probability that X lies is:
\[ P(\text{X lies}) = 1 - P(\text{X speaks truth}) = 1 - 0.6 = 0.4 = \frac{4}{10} \]

2. **Determine Y's probabilities:**
- Y speaks the truth in 50% of the cases, which can be expressed as:
\[ P(\text{Y speaks truth}) = 0.5 = \frac{5}{10} \]
- Therefore, the probability that Y lies is:
\[ P(\text{Y lies}) = 1 - P(\text{Y speaks truth}) = 1 - 0.5 = 0.5 = \frac{5}{10} \]

3. **Identify the cases where they contradict each other:**
- Case 1: X speaks the truth and Y lies.
- Case 2: X lies and Y speaks the truth.

4. **Calculate the probabilities for each case:**
- For Case 1 (X speaks truth, Y lies):
\[ P(\text{X speaks truth and Y lies}) = P(\text{X speaks truth}) \times P(\text{Y lies}) = \frac{6}{10} \times \frac{5}{10} = \frac{30}{100} = \frac{3}{10} \]

- For Case 2 (X lies, Y speaks truth):
\[ P(\text{X lies and Y speaks truth}) = P(\text{X lies}) \times P(\text{Y speaks truth}) = \frac{4}{10} \times \frac{5}{10} = \frac{20}{100} = \frac{2}{10} \]

5. **Add the probabilities of both cases to find the total probability of contradiction:**
\[ P(\text{Contradiction}) = P(\text{Case 1}) + P(\text{Case 2}) = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2} \]

### Final Answer:
The probability that X and Y contradict each other is:
\[ \frac{1}{2} \]