A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is:
Answer: 23116 (Option C)
Solutions · 1
Official solution
Since there are 6 blue and 6 green balls, a total of 6 pairs are drawn. For every pair to have one ball of each colour, we need a perfect matching between the 6 blue and 6 green balls.
The total number of ways to partition 12 distinguishable balls into 6 ordered pairs is (2!)66!12!×6! (accounting for the sequence of draws), and the number of "good" pairings (blue matched with green) is 6!×6! (each of the 6! ways to pair up blue with green balls, times 6! orders in which the pairs can be drawn).
\[P = \frac{6!\times6!}{\dfrac{12!}{(2!)^6}} = \frac{16}{231}\]
*Note: the exact option lettering (A/B/D) could not be reliably recovered from the source; the computed and confirmed answer is 23116 (given as option C in the source).*