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Question

$$A$$ speaks truth in $$60\%$$ cases and $$B$$ in $$90\%$$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?


A
0.42
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B
0.58
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C
0.46
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D
None of these
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Solution

The correct option is A $$0.42$$
Let $$E=$$ The event of $$A$$ speaking truth and 
$$F=$$ event of $$B$$ speaking truth 

$$ \displaystyle \therefore P\left( E \right) =\frac { 60 }{ 100 } =\frac { 6 }{ 10 } $$ and $$ \displaystyle P\left( F \right) =\frac { 90 }{ 100 } =\frac { 9 }{ 10 }  $$

Probability of $$A$$ and $$B$$ contradicting each other $$P(E\overline { F } $$ or $$\overline { F } E$$)
$$P\left( E\overline { F }  \right) +P\left( \overline { E } F \right) =P\left( E \right) P\left( \overline { F }  \right) +P\left( \overline { E }  \right) P\left( F \right)  $$
$$=P\left( E \right) \left( 1-P\left( F \right)  \right) +\left( 1-P\left( E \right)  \right) P\left( F \right)  $$
$$ \displaystyle =\frac { 6 }{ 10 } \left( 1-\frac { 9 }{ 10 }  \right) +\left( 1-\frac { 6 }{ 10 }  \right) \frac { 9 }{ 10 } $$
$$ \displaystyle =\frac { 6 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 4 }{ 10 } \times \frac { 9 }{ 10 } =\frac { 42 }{ 100 }  $$
$$\therefore A$$ and $$B$$ are likely to contradicts each other in $$42\%$$ cases.

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