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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    The probability that atleast one of the events A and B occurs, is 0.6. If A and B occur simultaneously with probability 0.2, then \[p(\bar{A})+P(\bar{B})\] is equal to

    A)  0.4                          

    B) 0.8    

    C) 1.2  

    D) 1.4

    Correct Answer: C

    Solution :

    We have, \[P(A\cup B)=0.6,P(A\cap B)=0.2\] \[\therefore \]  \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[\Rightarrow \]               \[0.6=P(A)+P(B)-0.2\] \[\Rightarrow \]               \[P(A)+P(B)=0.8\] \[\Rightarrow \]               \[1-P(\bar{A})+1-P(\bar{B})=0.8\] \[\Rightarrow \]               \[P(\bar{A})+P(\bar{B})=1.2\]                                


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