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
B
0.58
C
0.46
D
None of these

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.Maths

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