JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    Three numbers are chosen at random without replacement from 1, 2, 3, ..., 10. The probability that the minimum of the chosen numbers is 4 or their maximum is 8 is

    A) \[\frac{11}{40}\]                       

    B) \[\frac{3}{10}\]

    C) \[\frac{1}{40}\]             

    D)  none of these.

    Correct Answer: A

    Solution :

    [a] : Let us define the events in the following way: A : 4 being the minimum number B : 8 being the maximum number \[A\cap B:1\] being the minimum number and 8 being the maximum number Therefore \[P(A)=\frac{^{6}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{15}{120}\] \[P(B)=\frac{^{7}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{21}{120}\]and \[P(A\cap B)=\frac{^{3}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{3}{120}\] \[\therefore \]The required probability, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[=\frac{15}{120}+\frac{21}{120}-\frac{3}{120}=\frac{33}{120}=\frac{11}{40}\]

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