Probability of Drawing Mixed Color Pairs Without Replacement

To find the probability that each of the $6$ drawn pairs contains one blue ball and one green ball, we can calculate the total number of ways to partition the $12$ balls into $6$ pairs and the number of favorable ways to form such pairs.

Method 1: Sequential Probability Approach

Let us draw pairs sequentially without replacement:

1. First Pair ($12$ balls remaining: $6$ blue, $6$ green):

   • Total ways to choose $2$ balls from $12$: $\binom{12}{2} = 66$
   • Favorable ways to choose $1$ blue and $1$ green: $\binom{6}{1} \times \binom{6}{1} = 36$
   • Probability $P_1 = \frac{36}{66} = \frac{6}{11}$

2. Second Pair ($10$ balls remaining: $5$ blue, $5$ green):

   • Total ways to choose $2$ balls from $10$: $\binom{10}{2} = 45$
   • Favorable ways to choose $1$ blue and $1$ green: $\binom{5}{1} \times \binom{5}{1} = 25$
   • Probability $P_2 = \frac{25}{45} = \frac{5}{9}$

3. Third Pair ($8$ balls remaining: $4$ blue, $4$ green):

   • Total ways to choose $2$ balls from $8$: $\binom{8}{2} = 28$
   • Favorable ways to choose $1$ blue and $1$ green: $\binom{4}{1} \times \binom{4}{1} = 16$
   • Probability $P_3 = \frac{16}{28} = \frac{4}{7}$

4. Fourth Pair ($6$ balls remaining: $3$ blue, $3$ green):

   • Total ways to choose $2$ balls from $6$: $\binom{6}{2} = 15$
   • Favorable ways to choose $1$ blue and $1$ green: $\binom{3}{1} \times \binom{3}{1} = 9$
   • Probability $P_4 = \frac{9}{15} = \frac{3}{5}$

5. Fifth Pair ($4$ balls remaining: $2$ blue, $2$ green):

   • Total ways to choose $2$ balls from $4$: $\binom{4}{2} = 6$
   • Favorable ways to choose $1$ blue and $1$ green: $\binom{2}{1} \times \binom{2}{1} = 4$
   • Probability $P_5 = \frac{4}{6} = \frac{2}{3}$

6. Sixth Pair ($2$ balls remaining: $1$ blue, $1$ green):

   • Probability $P_6 = \frac{1}{1} = 1$

The required total probability $P$ is the product of all these sequential probabilities: $P = P_1 \times P_2 \times P_3 \times P_4 \times P_5 \times P_6$

$P = \left(\frac{6}{11}\right) \times \left(\frac{5}{9}\right) \times \left(\frac{4}{7}\right) \times \left(\frac{3}{5}\right) \times \left(\frac{2}{3}\right) \times 1$

Simplifying the expression: $P = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{11 \times 9 \times 7 \times 5 \times 3 \times 1} = \frac{720}{10395}$

Dividing the numerator and denominator by $45$: $P = \frac{16}{231}$

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Method 2: Combinatorial Partition Approach

1. Total Number of Ways ($N_{\text{total}}$): The number of ways to divide $12$ distinct balls into $6$ unordered pairs of size $2$ is given by: $N_{\text{total}} = \frac{12!}{2^6 \cdot 6!} = 11 \times 9 \times 7 \times 5 \times 3 \times 1 = 10395$

2. Favorable Number of Ways ($N_{\text{favorable}}$): To ensure every pair consists of $1$ blue ball and $1$ green ball, we match each of the $6$ distinct blue balls with a distinct green ball. The number of such pairings is equivalent to the number of permutations of $6$ green balls for the $6$ blue balls: $N_{\text{favorable}} = 6! = 720$

3. Required Probability: $P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{720}{10395} = \frac{16}{231}$

Thus, the probability that each drawn pair consists of one blue and one green ball is $\frac{16}{231}$.