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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    Let \[0<P(A)<1,0<P(B)<1\]and \[P(A\cup B)=P(A)+P(B)-P(A).P(B),\]then

    A)  \[P(B/A)=P(B)-P(A)\]

    B)  \[P(A'-B')=P(A')-P(B')\]

    C)  \[P(A\cup B)'=P(A).P(B)'\]

    D)  \[P(A/B)=P(A)-P(B)\]

    Correct Answer: C

    Solution :

    Since, \[P(A\cap B)=P(A).P(B)\] It means A and B are independent events, so A' and B' are also independent. \[\therefore \]\[P(A\cup B)'=P(A'\cap B')\] \[=P(A').P(B')\] Alternative \[P(A\cup B)'=1-P(A\cup B)\] \[=1-[P(A)+P(B)-P(A).P(B)]\] \[=[1-P(A)][1-P(B)]\] \[=P(A)'.P(B)'\]


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