Question

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

Answer 1

Hint: We solve this problem by starting with changing the percentage of cases to probability using the formula for the probability of getting an event A, $P\left( A \right)=\dfrac{n}{N}$, where n is occurrences of event A and total possible outcomes are N. Then we find the probability that both contradict each other. Then we find the percentage of cases they contradict using this probability. Then we find the probability that they are telling the truth given contradiction occurs and compare them.

Complete step-by-step solution:
Let us consider the event of A telling the truth as M and B telling the truth as N.
Now let us find the probability of A telling truth. We are given that A speaks the truth in 60% of the cases, that is in 100 cases A tells the truth in 60 cases.
So, let us consider the formula for probability
If a random experiment can result in N number of cases and n of them are favourable to the occurrence of event A, then the probability of occurrence of A is given by,
$P\left( A \right)=\dfrac{n}{N}$
So, using the above formula we can calculate the probability as
$\begin{align}
  & \Rightarrow P\left( M \right)=\dfrac{60}{100} \\
 & \Rightarrow P\left( M \right)=\dfrac{3}{5} \\
\end{align}$
Then probability of A telling lies is
$\begin{align}
  & \Rightarrow P\left( {{M}^{C}} \right)=1-P\left( M \right) \\
 & \Rightarrow P\left( {{M}^{C}} \right)=1-\dfrac{3}{5} \\
 & \Rightarrow P\left( {{M}^{C}} \right)=\dfrac{2}{5} \\
\end{align}$
Similarly let us find the probability of B telling the truth. We are given that B speaks truth in 90% of the cases, that is in 100 cases A tells the truth in 90 cases.
So, using the above formula we can calculate the probability as
$\begin{align}
  & \Rightarrow P\left( N \right)=\dfrac{90}{100} \\
 & \Rightarrow P\left( N \right)=\dfrac{9}{10} \\
\end{align}$
Then probability of B telling lies is
$\begin{align}
  & \Rightarrow P\left( {{N}^{C}} \right)=1-P\left( N \right) \\
 & \Rightarrow P\left( {{N}^{C}} \right)=1-\dfrac{9}{10} \\
 & \Rightarrow P\left( {{N}^{C}} \right)=\dfrac{1}{10} \\
\end{align}$
Now we need to find the percent of cases when both contradict each other. They contradict each other when one is telling the truth and other is telling the lie.
So, our required probability will be
$\begin{align}
  & \Rightarrow P\left( M \right)P\left( {{N}^{C}} \right)+P\left( {{M}^{C}} \right)P\left( N \right) \\
 & \Rightarrow \left( \dfrac{3}{5} \right)\left( \dfrac{1}{10} \right)+\left( \dfrac{2}{5} \right)\left( \dfrac{3}{5} \right) \\
 & \Rightarrow \dfrac{3}{50}+\dfrac{6}{25} \\
 & \Rightarrow \dfrac{15}{50} \\
 & \Rightarrow \dfrac{3}{10} \\
\end{align}$
So, the probability of A and B contradicting each other is $\dfrac{3}{10}$.
As we need it in percent let us find the number of cases, they contradict out of 100 cases.
$\begin{align}
  & \Rightarrow \dfrac{3}{10}\times 100 \\
 & \Rightarrow 30 \\
\end{align}$
So, they contradict each other in 30 cases among 100 cases, that is they contradict each other in 30% of the cases over the same fact.
To find who has more weightage in telling the truth while they are contradicting each other, we need to find the probability of A or probability of B telling the truth given that they are contradicting each other.
Let us consider the formula for Bayes Theorem.
$P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$
Using this formula, let us find the probability of A telling the truth given that A and B contradict each other.
$\begin{align}
  & \Rightarrow \dfrac{P\left( A\text{ telling the truth during contradiction} \right)}{P\left( A\text{ and }B\text{ contradicting each other} \right)} \\
 & \Rightarrow \dfrac{P\left( M \right)P\left( {{N}^{C}} \right)}{\dfrac{3}{10}} \\
 & \Rightarrow \dfrac{\dfrac{3}{5}\times \dfrac{1}{10}}{\dfrac{3}{10}} \\
 & \Rightarrow \dfrac{\dfrac{3}{50}}{\dfrac{3}{10}} \\
 & \Rightarrow \dfrac{10}{50} \\
 & \Rightarrow \dfrac{1}{5} \\
\end{align}$
Then the probability of B telling the truth during the contradiction is
$\begin{align}
  & \Rightarrow 1-P\left( A\text{ telling truth given there is a contradiction} \right) \\
 & \Rightarrow 1-\dfrac{1}{5} \\
 & \Rightarrow \dfrac{4}{5} \\
\end{align}$
So, we can see that probability of B telling truth is greater than probability of A telling the truth in the cases of contradiction.
So, we can say that in the cases of contradiction, the statement of B carries more weight than A as he speaks truth in more number of cases.

Note: In this question, one might make a mistake by taking the formula for probability given above as $P\left( A \right)=\dfrac{N}{n}$. But it wrong as N should be in the denominator. And another possibility of mistake is while applying the formula for Bayes, one might forget to take the numerator as $P\left( M \right)P\left( {{N}^{C}} \right)$ and take it as $P\left( M \right)$. But as we are calculating the probability while contradiction, we need to multiply it with $P\left( {{N}^{C}} \right)$.