Question

# Given two independent events $$A$$ and $$B$$ such that $$P\left( A \right) =0.3,P\left( B \right) =0.6$$. Find(i) $$P\left( A \,and \,B \right)$$(ii) $$P\left( A \,and \,not \,B \right)$$(iii) $$P\left( A \,or \,B \right)$$(iv) $$P\left( neither \,A \,nor \,B \right)$$If the sum of the above probablities is $$m$$ enter $$100m$$

Solution

## It is given that $$P\left( A \right) =0.3$$ and $$P\left( B \right) =0.6$$Also, $$A$$ and $$B$$ are independent events.(i) $$\therefore P\left( A and B \right) =P\left( A \right) \cdot P\left( B \right)$$$$\Rightarrow P\left( A\cap B \right) =0.3\times 0.6=0.18$$(ii) $$P\left( A and not B \right) =P\left( A\cap B\prime \right)$$$$=P\left( A \right) -P\left( A\cap B \right)$$$$=0.3-0.18$$$$=0.12$$(iii) $$P\left( A or B \right) = P\left( A\cup B \right)$$$$=P\left( A \right) +P\left( B \right) -P\left( A\cap B \right)$$$$=0.3+0.6-0.18$$$$=0.72$$(iv) $$P\left( neither A nor B \right) = P\left( A\prime \cap B\prime \right)$$$$=P\left( \left( A\cup B \right) \prime \right)$$$$=1-P\left( A\cup B \right)$$$$=1-0.72$$$$=0.28$$MathematicsRS AgarwalStandard XII

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