Question

A speaks truth in 60% of the cases, while B in 90% of cases. In what percent of cases they are likely to contradict each other in stating the same fact? In the case of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?

Answer 1

Hint: Find the probability of A speaks truth and of B speaks truth then find A does not speak truth and of B does not speak truth after this consider two cases and find their probabilities.

Complete step-by-step answer:
We know that probability of not an event is computed by subtracting the probability of event from one.
Suppose A is the event that person A speaks truth and B is the event that person B speaks truth.
 Here in the question we know that person A speaks truth in 60% cases, thus the probability that person A speaks truth is \[P\left( A \right)\] which is computed as,
 \[
  P\left( A \right) = 60\% \\
  P\left( A \right) = \dfrac{{60}}{{100}} \\
  P\left( A \right) = \dfrac{3}{5} \\
 \]
We also know that person B speaks truth in 90% cases, thus the probability that person B speaks truth is \[P\left( B \right)\] which is computed as,
\[
  P\left( B \right) = 90\% \\
  P\left( B \right) = \dfrac{{90}}{{100}} \\
  P\left( B \right) = \dfrac{9}{{10}} \\
 \]
Now we can find the probability that A does not speak truth by subtracting the probability of A speaking truth from one.
Finding the probability that A does not speaks truth \[P\left( {A'} \right)\],
\[
  P\left( {A'} \right) = 1 - P\left( A \right) \\
  P\left( {A'} \right) = 1 - \dfrac{3}{5} \\
  P\left( {A'} \right) = \dfrac{{5 - 3}}{5} \\
  P\left( {A'} \right) = \dfrac{2}{5} \\
 \]
Thus, the probability that A does not speak truth is \[P\left( {A'} \right) = \dfrac{2}{5}\].
Now we can find the probability that B does not speak truth by subtracting the probability of B speaking truth from one.
Finding the probability that B does not speaks truth \[P\left( {B'} \right)\],
\[
  P\left( {B'} \right) = 1 - P\left( B \right) \\
  P\left( {B'} \right) = 1 - \dfrac{9}{{10}} \\
  P\left( {B'} \right) = \dfrac{{10 - 9}}{{10}} \\
  P\left( {B'} \right) = \dfrac{1}{{10}} \\
 \]
Thus, the probability that B does not speak the truth is \[P\left( {B'} \right) = \dfrac{1}{{10}}\].
Now we have all the required probabilities.
Now we will consider case one where A speaks truth and B is not.
Now we will find the probability for case one and for this we will multiply the probability of A speaks truth by the probability of B does not speaks truth,
\[
  P\left( {{\text{case }}{\text{1}}} \right) = P\left( A \right)P\left( {B'} \right) \\
  P\left( {{\text{case }}{\text{1}}} \right) = \dfrac{3}{5} \cdot \dfrac{1}{{10}} \\
  P\left( {{\text{case }}{\text{1}}} \right) = \dfrac{3}{{50}} \\
 \]
Thus, the probability that A speaks truth but B does not is \[\dfrac{3}{{50}}\].
Simillary, we will consider case two where A does not speak truth and B speaks truth.
Now we will find the probability for case two, for this we will multiply the probability of A does not speaks truth by the probability of B speaks truth,
\[
  P\left( {{\text{case 2}}} \right) = P\left( {A'} \right)P\left( B \right) \\
  P\left( {{\text{case }}2} \right) = \dfrac{2}{5} \cdot \dfrac{9}{{10}} \\
  P\left( {{\text{case }}2} \right) = \dfrac{9}{{25}} \\
 \]
Thus, the probability that A does not speak truth and B speaks truth is \[\dfrac{9}{{25}}\].
Now we will add the probability of case one to of case two and multiply the result by to find the percent of cases they likely to contradict each other,
\[
  \left( {\dfrac{3}{{50}} + \dfrac{9}{{25}}} \right) \times 100\% = \left( {\dfrac{3}{{50}} + \dfrac{9}{{25}} \cdot \dfrac{2}{2}} \right) \times 100\% \\
   = \left( {\dfrac{3}{{50}} + \dfrac{{18}}{{50}}} \right) \times 100\% \\
   = \left( {\dfrac{{3 + 18}}{{50}}} \right) \times 100\% \\
   = \left( {\dfrac{{21}}{{50}}} \right) \times 100\% \\
   = \left( {21} \right) \times 2\% \\
   = 42\% \\
 \]
Thus, In 42% cases both the people are likely to contradict each other.
As, we consider first case that “A speaks truth and B is not” and got the probability \[\dfrac{3}{{50}}\] so, in the case of contradiction statement of B have more weightage when the probability of B saying correct and A is saying wrong is more.

Note:
Probability of event A and event B always multiplies while finding the probability of event A and event B. Probability of event A and event B is always adds while finding the probability of event A or event B.s