Question

A speaks truth in 75% of cases and B in 80% of cases. In what percent of cases are they likely to contradict each other in narrating the same event?
(A) 20%
(B) 25%
(C) 30%
(D) 35%

Answer 1

Hint: The contradiction between the two will occur if one of them speaks the truth and the other one lies. Form two different cases for their lying separately and then use the rule of probability of mutually exclusive events to calculate the overall probability.

Complete step by step answer:
According to the question, A speaks truth in 75% of cases and B in 80% of cases. If $P\left( A \right)$ and $P\left( B \right)$are probabilities of A and B speaking truth respectively, then:
$P\left( A \right) = \dfrac{{75}}{{100}} = \dfrac{3}{4},P\left( B \right) = \dfrac{{80}}{{100}} = \dfrac{4}{5}$
$P\left( {\overline A } \right)$ and $P\left( {\overline B } \right)$ will be their respective probabilities of speaking false. So, we have:
$P\left( {\overline A } \right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}{\text{ and }}P\left( {\overline B } \right) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$
Now, the contradiction between them can occur in 2 ways. We will solve both the cases separately.
Case 1: A speaks the truth but B lies. The probability of this case will be:
$
   \Rightarrow {P_1} = P\left( A \right) \times P\left( {\overline B } \right), \\
   \Rightarrow {P_1} = \dfrac{3}{4} \times \dfrac{1}{5}, \\
   \Rightarrow {P_1} = \dfrac{3}{{20}} \\
$
Case 2: A tells a lie and B speaks the truth. The probability of this case will be:
$
   \Rightarrow {P_2} = P\left( {\overline A } \right) \times P\left( B \right), \\
   \Rightarrow {P_2} = \dfrac{1}{4} \times \dfrac{4}{5}, \\
   \Rightarrow {P_2} = \dfrac{4}{{20}} \\
$
Since, both the cases are mutually exclusive, the overall probability will be the sum of the probabilities of two cases:
$
   \Rightarrow P = {P_1} + {P_2}, \\
   \Rightarrow P = \dfrac{3}{{20}} + \dfrac{4}{{20}}, \\
   \Rightarrow P = \dfrac{7}{{20}} \\
$
Percentage occurrence of contradiction $ = \dfrac{7}{{20}} \times 100\% = 35\% $
Therefore, in 35% of the cases A and B are likely to contradict each other. (D) option is correct.

Note: If two events A and B with the probability of their occurrence $P\left( A \right)$ and $P\left( B \right)$ are mutually exclusive, then the probability of occurrence of either of them is:
$ \Rightarrow P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$