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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    Let X and Y are two events such that\[P\left( X\cup Y \right)=P(X\cap Y).\] Statement1:\[P(X\cap Y')=P\left( X'\cap Y \right)=0\] Statement 2; \[P(X)+P=2P\left( X\cap Y \right)\]   JEE Main Online Paper (Held On 07 May 2012)

    A) Statement 1 is false. Statement 2 is true. Statement 1 is true. Statement 2 is true,

    B)                        Statement 2 is not a correct explanation of Statement 1.

    C)                         Statement 1 is true. Statement 2 is false.

    D)                        Statement 1 is true. Statement 2 is true; Statement 2 is a correct explanation of Statement 1.      

    Correct Answer: B

    Solution :

                    Let X and V be two events such that \[P\left( X\cup Y \right)=P\left( X\cap Y \right)\]                                               ?(1) We know \[P\left( X\cup Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)\] \[P\left( X\cap Y \right)=P\left( X \right)+P\left( Y \right)-P\left( X\cap Y \right)\] (from (1) \[\Rightarrow \]\[P\left( X \right)+P\left( Y \right)=2P\left( X\cap Y \right)\] Hence, Statement - 2 is true. Now, \[P\left( X\cap Y' \right)=P\left( X \right)-P\left( X\cap Y \right)\] and\[P\left( X'\cap Y \right)=P\left( Y \right)-P\left( X\cap Y \right)\] This implies statement-1 is also true.


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