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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The probability distribution of a random variable X is
    \[x=x\] \[-1\] \[0\] \[1\] \[2\]
    \[P(X=X)\] \[0.2\] \[0.1\] \[0.4\] \[\lambda \]
    Then, \[P(0<X\le 2)\] is equal to

    A)  \[0.8\]              

    B)  \[0.6\]

    C)  \[0.5\]              

    D)  \[0.7\]

    Correct Answer: D

    Solution :

    We know that, total probability \[=1\] \[\Rightarrow \] \[P(-1)+P(1)+P(2)=1\] \[\Rightarrow \] \[0.2+0.1+0.4+\lambda =1\] \[\Rightarrow \] \[\lambda =1-0.7=0.3\] Then \[P(0<x\le 2)=P(1)+P(2)\] \[=0.4+0.3\] \[=0.7\]


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