Question

If $$A$$ and $$B$$ are two events such that $$P\left( A \right) = \displaystyle\frac { 1 }{ 4 } , P\left( B \right) = \displaystyle\frac { 1 }{ 2 }$$ and $$P\left( A\cap B \right) = \displaystyle\frac { 1 }{ 8 }$$, find $$8P\left( \bar A \ and\ \bar{B} \right)$$

Solution

It is given that, $$P\left( A \right) =\displaystyle\frac { 1 }{ 2 }$$ and $$P\left( A\cap B \right) = \displaystyle\frac { 1 }{ 8 }$$$$P\left( not on A and not on B \right) =P\left( A\prime \cap B\prime \right)$$$$P\left( not on A and not on B \right) = P\left( \left( A\cup B \right) \right) \prime \left[ A\prime \cap B\prime = \left( A\cup B \right) \prime \right]$$$$=1 - P\left( A\cup B \right)$$$$=1 - \left[ P\left( A \right) + P\left( B \right) - P\left( A\cap B \right) \right]$$$$= 1 - \left[ \displaystyle\frac { 1 }{ 4 } + \displaystyle\frac { 1 }{ 2 } - \displaystyle\frac { 1 }{ 8 } \right]$$$$= 1 - \displaystyle\frac { 5 }{ 8 }$$$$= \displaystyle\frac { 3 }{ 8 }=0.27$$MathematicsRS AgarwalStandard XII

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