A) 36%
B) 42%
C) 48%
D) None of these
Correct Answer: B
Solution :
[b] Let E= the event of A's speaking truth F = the event of B's speaking truth \[\therefore P(E)=\frac{60}{100}=\frac{3}{5}\] \[\therefore P(\overline{E})=1-P(E)=1-\frac{3}{5}=\frac{2}{5}\] Again \[P(F)=\frac{90}{100}=\frac{9}{10}\] \[\therefore P(F)=\frac{1}{10}\] A and B will contradict each other when A speak truth and B tells a lies and vice-versa. They are mutually exclusive events, i.e. Case-I: When A speaks truth and B tells a lies means \[E\cap \overline{F}.\] Case-II: When A speaks tells a lies and B speaks truth means \[\overline{E}\cap F.\] Thus, the required probability \[=P\left( E\cap \overline{F} \right)+P\left( \overline{E}\cap F \right)=P\left( E \right).P\left( \overline{F} \right)+P\left( \overline{E} \right).P\left( F \right)\]\[=\frac{3}{5}\times \frac{1}{10}+\frac{2}{5}\times \frac{9}{10}=\frac{3}{10}+\frac{18}{50}=\frac{21}{50}\] So, the percentage to contradict each other \[=\frac{21}{50}\times 100%=42%\] Hence, option [b] is correct.
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