ProbabilityMedium
Question
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 :
Options
A.$\frac{63}{925}$
B.$\frac{17}{231}$
C.$\frac{16}{231}$
D.$\frac{64}{925}$
Solution
Sol. [Key Concept: Count pairings of the $12$ balls into $6$ pairs where every pair is one blue and one green: probability $= \frac{6!\ 6!{\ 2}^{6}}{12!} \cdot (arrangements) = \frac{16}{231}$.]
Probability $= \frac{\underline{6} \times \underline{6}}{\left( \frac{\underline{12}}{(\check{2})^{6}\lfloor 6} \right) \times \mid 6} = \frac{16}{231}$
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