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  • Question
  • An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.


  • Options
  • A. 5/204
  • B. 1/204
  • C. 13/204
  • D. None of these

  • Correct Answer
  • 1/204 

    Explanation

    Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

     Required probability =  P A ? B ? C  

    P A   P B A   P C A ? B

     Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

    When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

      ? P B A = 3 17  

    Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

      ? P C A ? B   = 2 16 = 1 8

     Hence the required probability =  2 9 × 3 17 × 1 8 = 1 204


  • More questions

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    • 2. 
      In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten's place. If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is

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