2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    Let A and B be two mutually exclusive events such    that    \[P(A\cap {{B}^{c}})=0.25\]and \[P({{A}^{c}}\cap B)=0.5\]. Then, \[P\{(A\cup {{B}^{c}})\}\] is equal to

    A)  \[0.25\]             

    B)  \[0.50\]

    C)  \[0.75\]             

    D)  \[0.40\]

    Correct Answer: A

    Solution :

    Given, \[P(A\cap {{B}^{c}})=0.25\] and \[P({{A}^{c}}\cap B)=0.5\] \[\therefore \] \[P\{{{(A\cup B)}^{c}}\}=1-P(A\cup B)\] \[=1-\{P(A\cap {{B}^{c}})+P({{A}^{c}}\cap B)+P(A\cap B)\}\] \[=1-(0.25+0.5+0)\] [ \[\because \] A and B are mutually exclusive events \[P(A\cap B)=0\]] \[=1-0.75=0.25\]


You need to login to perform this action.
You will be redirected in 3 sec spinner