JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer A random variable X has the probability distribution
    \[x\] 1 2 3 4 5 6 7 8
    \[p(x)\] 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
    For the events E = {X is a prime number} and\[F=\{X<4\},\]the probability\[P(E\cup F)\]is

    A) 0.87  

    B)                       0.77       

    C)        0.35       

    D)        0.50

    Correct Answer: B

    Solution :

    If A and B are two events associated with a random experiment. Then, \[P(A\cup B)=P(A)+P(B)-P(A\cap B),\] which is equal to the atleast one of the event is occur. E = {X is a prime number} = { 2, 3, 5, 7} \[P(E)=P(X=2)+P(X=3)\]                                 \[+P(X=5)+P(X=7)\] \[\Rightarrow \]\[P(E)=0.23+0.12+0.20+0.07+0.62\] and \[F=\{X<4\}=\{1,2,3\}\] \[P(F)=P(X=1)+P(X=2)+P(X=3)\] \[\Rightarrow \] \[P(F)=0.15+0.23+0.12=0.5\] \[E\cap F=\]{X is prime number as well as < 4} = {2, 3} \[P(E\cap F)=P(X=2)+P(X=3)\] \[=0.23+0.12=0.35\] \[\therefore \]Required probability, \[P(E\cup F)=P(E)+P(F)-P(E\cap F)\] \[P(E\cup F)=0.62+0.5-0.35\] \[P(E\cup F)=0.77\]


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