Question

P speaks truth in 70% of the cases and Q in 30% of the cases. In what percent of cases are they likely to agree in stating the same fact? Do you think, when they agree, means both are speaking the truth ?

Answer 1

Hint: Here, we have to consider two events, where A: P speaks the truth and B: Q speaks the truth. We have to apply the formula that probability of P and Q stating the same fact,
$P(S)=P(A)\cap P(\bar{B})+P(B)\cap P\left( {\bar{A}} \right)$, for independent events use the formula :
$P(S)=P(A)P\left( {\bar{B}} \right)+P(B)P\left( {\bar{A}} \right)$

Complete step-by-step solution -
Here, we have to consider two events. Let $A$ be the event that P speaks the truth and let $B$be the event that Q speaks the truth.
Similarly, let $\bar{A}$ be the event that P speaks lie and $\bar{B}$ be the event that Q speaks lie.
We are given that P speaks truth in 70% of the cases. i.e. probability of P speaking the truth is:
$\begin{align}
  & P(A)=\dfrac{70}{100} \\
 & P(A)=\dfrac{7}{10} \\
\end{align}$
Also, Q speaks truth in 30% 0f the cases. i.e. probability of Q speaking the truth is:
$\begin{align}
  & P(B)=\dfrac{30}{100} \\
 & P(B)=\dfrac{3}{10} \\
\end{align}$
We can also say that probability of P speaking the lie is:
$\begin{align}
  & P\left( {\bar{A}} \right)=1-P(A) \\
 & P\left( {\bar{A}} \right)=1-\dfrac{7}{10} \\
\end{align}$
By taking the LCM and cross multiplying we get:
$\begin{align}
  & P\left( {\bar{A}} \right)=\dfrac{10-7}{10} \\
 & P\left( {\bar{A}} \right)=\dfrac{3}{10} \\
\end{align}$
Similarly, probability of Q speaking the lie is:
$\begin{align}
  & P\left( {\bar{B}} \right)=1-P(B) \\
 & P\left( {\bar{B}} \right)=1-\dfrac{3}{10} \\
\end{align}$
By taking LCM and cross multiplying we get:
$\begin{align}
  & P\left( {\bar{B}} \right)=\dfrac{10-3}{10} \\
 & P\left( {\bar{B}} \right)=\dfrac{7}{10} \\
\end{align}$
Let $S$ be the event that both P and Q are likely to agree in the same fact. i.e. contradict with each other.
Both contradict with each other = [A true and B lies] or [B true and A lies]
Now, probability of P and Q contradict with each other is:
$P(S)=P(A)\cap P(\bar{B})+P(B)\cap P\left( {\bar{A}} \right)$
Here, all the events are independent, therefore we can write:
$\begin{align}
  & P(S)=P(A)P\left( {\bar{B}} \right)+P(B)P\left( {\bar{A}} \right) \\
 & P(S)=\dfrac{7}{10}\times \dfrac{7}{10}+\dfrac{3}{10}\times \dfrac{3}{10} \\
 & P(S)=\dfrac{49}{100}+\dfrac{9}{100} \\
\end{align}$
By taking the LCM and cross multiplying we get:
$\begin{align}
  & P(S)=\dfrac{49+9}{100} \\
 & P(S)=\dfrac{58}{100} \\
\end{align}$
By cancellation we obtain:
$P(S)=\dfrac{29}{50}$
The percentage of cases that they contradict with each other =$\dfrac{29}{50}\times 100=58%$

Note: Here, both P and Q likely agree in the same fact means both contradict with each other. So don’t get confused that both are agreeing in the same fact, which may lead to an incorrect formula.