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KVPY Sample Paper KVPY Stream-SX Model Paper-15

  • question_answer
    Let A, B, C be three events in a probability space. Suppose that \[P(A)=0.5,\text{ }P(B)=0.3,\text{ }P(C)=0.2,\]\[P(A\cap B)=0.15,\] \[P(A\cap C)=0.1\] and \[P(B\cap C)=0.06,\] the greatest possible value of \[P({{A}^{c}}\cap {{B}^{c}}\cap {{C}^{c}})\] is [Note: \[{{A}^{c}}\]denotes compliment of event A]

    A) 0.31     

    B) 0.25  

    C) 0

    D) 0.26

    Correct Answer: A

    Solution :

    \[P({{A}^{C}}\cap {{B}^{C}}\cap {{C}^{C}})=1-P(A)-P(B)-P(C)+\]\[P(A\cap B)+P(B\cap C)+P(C\cap A)-P(A\cap B\cap C)\] \[P{{({{A}^{C}}\cap {{B}^{C}}\cap {{C}^{C}})}_{max}}=1-P(A)-P(B)-P(C)\]\[+P(A\cap B)+P(B\cap C)+P(C\cap A)=0.31\]


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