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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) $$P\left( neither   A   nor   B \right) = P\left( A\prime \cap B\prime  \right) $$
$$=P\left( \left( A\cup B \right) \prime  \right) $$
$$=1-P\left( A\cup B \right) $$
$$=1-0.72$$
$$=0.28$$

Mathematics
RS Agarwal
Standard XII

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