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JEE Main & Advanced Mathematics Probability Question Bank Odds in favour and odds against Addition theorem on probability

  • question_answer
    A and B are two independent events. The probability that both A and B occur is \[\frac{1}{6}\] and the probability that neither of them occurs is \[\frac{1}{3}\]. Then the probability of the two events are respectively                                          [Roorkee 1989]

    A)                 \[\frac{1}{2}\]and \[\frac{1}{3}\]       

    B)                 \[\frac{1}{5}\]and \[\frac{1}{6}\]

    C)                 \[\frac{1}{2}\]and \[\frac{1}{6}\]       

    D)                 \[\frac{2}{3}\]and \[\frac{1}{4}\]

    Correct Answer: A

    Solution :

                       \[P(A\cap B)=P(A).P(B)=\frac{1}{6}\]                    \[P(\bar{A}\cap \bar{B})=\frac{1}{3}=1-P(A\cup B)\]                    \[\Rightarrow \frac{1}{3}=1-[P(A)+P(B)]+\frac{1}{6}\Rightarrow P(A)+P(B)=\frac{5}{6}.\]                                 Hence \[P(A)\] and \[P(B)\] are \[\frac{1}{2}\] and \[\frac{1}{3}.\]


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