QJEEM22Ma725b: JEE Main 2022 Mathematics - Class 12 JEEM22Ma725b Solution: Let E 1 and E 2 be two events such that the conditional…

$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$

$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$

$P({E_1} \cap {E_2}) = {1 \over 8}$

$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$

(A) $P({E_1} \cap {E_2}) = {1 \over 8}$ and $P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$\Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

(B) $P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$

$= 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$

$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$

$\Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$

(C) $P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$

$= {1 \over 6} - {1 \over 8} = {1 \over {24}}$

$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$\Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$

(D) $P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$

$= {1 \over 4} - {1 \over 8} = {1 \over 8}$

$P({E_1})P({E_2}) = {1 \over {24}}$

$\Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$