Multiple choice

A bag contains 10 black marbles, some red marbles and some green marbles. If the number of red marbles is 3 more than twice the number of green marbles and the total number of marbles in the bag is 31 (there are only three colour marbles in bag), then what is the probability of picking one red and one green marble when 2 marbles are drawn from the bag?

  1. 6/31

  2. 5/31

  3. 4/31

  4. None of these

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total = 31. Black = 10. Red + Green = 21. R = 2G + 3. (2G + 3) + G = 21 => 3G = 18 => G = 6, R = 15. Probability of one red and one green = (15C1 * 6C1) / 31C2 = (15 * 6) / (31 * 30 / 2) = 90 / 465 = 6/31.

AI explanation

Let the number of green marbles be g, which makes the number of red marbles 2g + 3, and solving the equation 10 + g + 2g + 3 = 31 gives g = 6, meaning there are 6 green and 15 red marbles. The total number of ways to pick 2 marbles from the 31 marbles is 31C2, which equals 465. The number of ways to pick one red and one green marble is 15 multiplied by 6, giving 90, so the probability is 90/465, which simplifies to 6/31.

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