Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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 :

  1. A $P({E_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})$
  2. B $P(E{'_1} \cap E{'_2}) = P(E{'_1})\,.\,P(E{_2})$
  3. C $P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$ Correct answer
  4. D $P(E{'_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})$

Solution

<p>$$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$$</p> <p>$$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$$</p> <p>$P({E_1} \cap {E_2}) = {1 \over 8}$</p> <p>$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$</p> <p>(A) $P({E_1} \cap {E_2}) = {1 \over 8}$ and $P({E_1})\,.\,P({E_2}) = {1 \over {24}}$</p> <p>$\Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$</p> <p>(B) $P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$</p> <p>$$ = 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$$</p> <p>$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$</p> <p>$\Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$</p> <p>(C) $P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$</p> <p>$= {1 \over 6} - {1 \over 8} = {1 \over {24}}$</p> <p>$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$</p> <p>$\Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$</p> <p>(D) $P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$</p> <p>$= {1 \over 4} - {1 \over 8} = {1 \over 8}$</p> <p>$P({E_1})P({E_2}) = {1 \over {24}}$</p> <p>$\Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$</p>

About this question

Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability

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