JEE MAIN - Mathematics (2022 - 25th June Morning Shift - No. 7)

Let E1 and E2 be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \over 2}$$, $$P({E_2}|{E_1}) = {3 \over 4}$$ and $$P({E_1} \cap {E_2}) = {1 \over 8}$$. Then :
$$P({E_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})$$
$$P(E{'_1} \cap E{'_2}) = P(E{'_1})\,.\,P(E{_2})$$
$$P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$$
$$P(E{'_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})$$

Explanation

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

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

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

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

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

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

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

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

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