Person A tells the truth 30% of the times and B tells the truth 40% of the times, independently. What is the minimum probability that they would contradict each other?

Explanation:

Let A: A tells the truth, A^c = A lies

B: B tells the truth, Bc = B lies 

So P(A) = 30% = \(\frac{30}{100}\) = \(\frac{3}{10}\)

and P(A^c) = 1 - \(\frac{3}{10}\) = \(\frac{7}{10}\)

Similarly, 

 P(B) = 40% = \(\frac{40}{100}\) = \(\frac{4}{10}\)

and P(Bc) = 1 - \(\frac{4}{10}\) = \(\frac{6}{10}\)

Therefore that the minimum probability that they would contradict each other i.e., one tell the truth and other lies

= P(A) × P(Bc) + P(B) × P(Ac)

 = \(\frac{3}{10}\) × \(\frac{6}{10}\) + \(\frac{7}{10}\) × \(\frac{4}{10}\) = \(\frac{18}{100}+\frac{28}{100}\) = \(\frac{46}{100}\) = 0.46

Option (3) is correct