A speaks the truth $\frac{5}{7}$ times and B speaks the truth $\frac{8}{9}$ times. What is the probability that they contradict each other in stating the same fact?

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**Solution:**

1. Let the probability that A speaks the truth be $P(A) = \frac{5}{7}$.

2. Then, the probability that A lies is $P(A') = 1 - P(A) = 1 - \frac{5}{7} = \frac{2}{7}$.

3. Let the probability that B speaks the truth be $P(B) = \frac{8}{9}$.

4. Then, the probability that B lies is $P(B') = 1 - P(B) = 1 - \frac{8}{9} = \frac{1}{9}$.

5. They contradict each other if one speaks the truth and the other lies. So, the required probability is:

$$P(	ext{contradiction}) = P(A) 	imes P(B') + P(A') 	imes P(B)$$

6. Substitute the values:

$$= \frac{5}{7} 	imes \frac{1}{9} + \frac{2}{7} 	imes \frac{8}{9}$$

$$= \frac{5}{63} + \frac{16}{63} = \frac{21}{63} = \frac{1}{3}$$

**Answer:** $\frac{1}{3}$

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