A speaks truth in 60% of the cases, while B in 40% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact?

Answer

To solve the problem of determining the percentage of cases in which A and B are likely to contradict each other in stating the same fact, we can follow these steps:

### Step 1: Define the probabilities
Let:
- \( P(E) \) = Probability that A speaks the truth = 60% = \( \frac{60}{100} = \frac{6}{10} \)
- \( P(F) \) = Probability that B speaks the truth = 40% = \( \frac{40}{100} = \frac{4}{10} \)

### Step 2: Calculate the probabilities of speaking falsely
- The probability that A does not speak the truth (A speaks falsely) is:
\[ P(E') = 1 - P(E) = 1 - \frac{6}{10} = \frac{4}{10} \]
- The probability that B does not speak the truth (B speaks falsely) is:
\[ P(F') = 1 - P(F) = 1 - \frac{4}{10} = \frac{6}{10} \]

### Step 3: Determine the probabilities of contradiction
A and B contradict each other if:
1. A speaks the truth and B speaks falsely, or
2. A speaks falsely and B speaks the truth.

Thus, we can express the probability of contradiction as:
\[ P(E \cap F') + P(E' \cap F) = P(E) \cdot P(F') + P(E') \cdot P(F) \]

### Step 4: Substitute the values
Substituting the values we calculated:
\[ P(E \cap F') = P(E) \cdot P(F') = \frac{6}{10} \cdot \frac{6}{10} = \frac{36}{100} \]
\[ P(E' \cap F) = P(E') \cdot P(F) = \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} \]

### Step 5: Add the probabilities of contradiction
Now, we can add these two probabilities:
\[ P(\text{Contradiction}) = P(E \cap F') + P(E' \cap F) = \frac{36}{100} + \frac{16}{100} = \frac{52}{100} \]

### Step 6: Convert to percentage
To express this as a percentage:
\[ \frac{52}{100} = 52\% \]

### Final Answer
Thus, A and B are likely to contradict each other in **52%** of cases.

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