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      Question

      A bag has ((x + 1)) red, ((x + 4)) green and ((x + 7))

      yellow balls. Find the probability of taking, without replacement, out 2 red, 2 green and 1 yellow ball if the probability of taking out 1 yellow ball at random from the bag is ((5/12)).
      A 45/476 Correct Answer Incorrect Answer
      B 47/476 Correct Answer Incorrect Answer
      C 51/476 Correct Answer Incorrect Answer
      D 55/476 Correct Answer Incorrect Answer
      E 59/476 Correct Answer Incorrect Answer

      Solution

      According to question: ((5/12) = (x + 7)C1/(3x + 12)C1) Or, ((5/12) = {(x + 7)/(3x + 12)}) Or, (15x + 60 = 12x + 84) Or, (3x = 24) Or, (x = 8) So, number of red balls = ((8 + 1) = 9) Number of green balls = ((8 + 4) = 12) And, number of yellow balls = ((8 + 7) = 15) Required probability = ((9C2 × 12C2 × 15C1)/36C5) = ({(36 × 66 × 15)/376992}) = (35640/376992) = (495/5236) = 45/476

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