📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!
  • google app store apple app store
    • ✖

      Question

      A box contains 5 red, 3 blue, and 2 green balls. Three

      balls are drawn at random without replacement.  What is the probability that  at least two of the drawn balls are red, given that at least one of the drawn balls is green?
      A 1/14 Correct Answer Incorrect Answer
      B 1/15 Correct Answer Incorrect Answer
      C 5/16 Correct Answer Incorrect Answer
      D 1/14 Correct Answer Incorrect Answer
      E 1/28 Correct Answer Incorrect Answer

      Solution

      Total balls = 10. P(at least 1 green) = 1 − P(no green) = 1 − [C(8,3) / C(10,3)] = 1 − 56/120 = 1 − 7/15 = 8/15 “At least 2 red and at least 1 green” ⇒ only possible composition in 3 draws: 2 red, 1 green P(2 red and 1 green) = [C(5,2)·C(2,1)] / C(10,3) = (10·2)/120 = 1/6 Required probability = P(≥2 red | ≥1 green) = (1/6) ÷ (8/15) = (1/6)·(15/8) = 15/48 = 5/16.

      Practice Next
      More Probability Questions

      Relevant for Exams:

      ask-question