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      Question

      A bag has ((x - 1)) red, ((x + 2)) green and ((x + 5))

      yellow balls. Find the probability of taking, without replacement, out 2 red, 1 green and 2 yellow balls if the probability of taking out 1 yellow ball at random from the bag is ((3/8)).
      A 8/79 Correct Answer Incorrect Answer
      B 9/79 Correct Answer Incorrect Answer
      C 10/79 Correct Answer Incorrect Answer
      D 12/79 Correct Answer Incorrect Answer
      E 15/79 Correct Answer Incorrect Answer

      Solution

      According to question: ((3/8) = (x + 5)C1/(3x + 6)C1) Or, ((3/8) = {(x + 5)/(3x + 6)}) Or, (9x + 18 = 8x + 40) Or, (x = 22) So, number of red balls = ((22 - 1) = 21) Number of green balls = ((22 + 2) = 24) And, number of yellow balls = ((22 + 5) = 27) Required probability = ((21C2 × 24C1 × 27C2)/72C5) = ({(210 × 24 × 351)/13991544}) = (1769040/13991544) = (70/553) = 10/79 

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