A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. The probability that 9th one examined is the last defective, is

A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. The probability that 9th one examined is the last defective, is

  1. A

     4C3×11C5 15C8

  2. B

     4C3×11C5 15C8×17

  3. C

     11C5 15C8×17

  4. D

     4C3 11C5×17

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    Solution:

    Let A and B be two events defined by

      A = Getting exactly 3 defectives in the examination of 8 record players. 

    = 9th record player is defective.

     Required probability =P(AB)=P(A)P(B/A)          …(i)

    Now,

           P(A)= 4C3×11C5 15C8

    and,

         P(B/A)=Probability that the 9th examined record player is defective given that there were 3 defective in the first 8 pieces examined

     P(B/A)=17

      Required probability = 4C3×11C5 15C8×17

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