A speaks truth in $75\%$ cases and B in $80\%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
(a) $70\%$ (b)$35\%$
(c) $25\%$ (d)$20\%$
what i have tried:
I would like to show about what i think, If QA is in Yes/No both A and B answer behave in similar to Exclusive-OR where agree$=0$ and disagree$=1$
$$\begin{array}{l|l|l}
\text{A} & \text{B} & \text{Fact} \\ \hline
\ T & T & \text{agree} \\ \hline
\ T & F & \text{disagree} \\ \hline
\ F & T & \text{disagree} \\ \hline
\ F & F & \text{agree}
\end{array}$$
Percentage of cases where they are likely to contradict each other=$80\%-75\%=5\%$
because till $75\%$ A speaks the truth and even B speaks truth
After $75\%$ A speaks false and B speaks truth and after $80\%$ both A & B speaks false
and so the answer is $5\%$ which is not there in option can any please help me this problem?
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$