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# For narrating some incidents A speaks truth in $75\%$ cases and B in $80\%$ of the cases. In narrating some cases both A and B are likely to contradict each other. What is the percentage of the cases in which they both likely contradict each other, narrating the same incident?(A) $25\%$(B) $40\%$(C) $35\%$(D) $10\%$

Last updated date: 22nd Jun 2024
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Hint:In this we apply the probability of happening an event and not happening of that particular event. So, the probability of A speaking truth is P (A) = $75\%$or $\dfrac{3}{4}$and not speaking truth or lie $P(\overline A )$ = $25\%$ or$\dfrac{1}{4}$. Similarly the probability of B speaking truth is P (B) = $80\%$ or $\dfrac{4}{5}$and not speaking truth is $P(\overline {B)}$= $20\%$ or$\dfrac{1}{5}$. In case of contradiction with each other is when A speaks truth and B speaks lie or A speaks lie and B speaks truth. For finding the probability we add both the situation and find the percentage.

Let A be Event that speaks the truth and B be Event that speaks the truth
So, the probability of A speaking truth P(A)= $75\%$
=$\dfrac{{75}}{{100}} = \dfrac{3}{4}$
The probability of B speaking truth P(B)= $80\%$
=$\dfrac{{80}}{{100}} = \dfrac{4}{5}$
We know $P(A)+P(\overline A )=1$ Hence,
Probability of A not speaking truth $P(\overline A )$= $1 - \dfrac{3}{4} = \dfrac{1}{4}$
Probability of B not speaking truth $P(\overline {B)}$ = $1 - \dfrac{4}{5}=\dfrac{1}{5}$
Now, A and B contradict each other=[A speaks truth and B speaks lie] or [A speaks lie and B speaks truth]
= $P(A) \times P(\overline B ) + P(\overline A ) \times P(B)$
=$(\dfrac{3}{4} \times \dfrac{1}{5}) + (\dfrac{1}{4} \times \dfrac{4}{5})$
=$\dfrac{3}{{20}} + \dfrac{4}{{20}}$
=$\dfrac{7}{{20}}$
For percentage we multiply $\dfrac{7}{{20}}$ by $100\%$
=$(\dfrac{7}{{20}} \times 100)\%$
=$35\%$
So, the case when A and B contradicts is $35\%$.
The correct answer is C=$35\%$.