A bag contains 6 blue and 6 green balls. Pairs of balls are drawn...

The probability that the first pair drawn consists of one blue and one green ball is:

$P_1 = \dfrac{^{6}C_{1} \times ^{6}C_{1}}{^{12}C_{2}} = \dfrac{36}{66} = \dfrac{6}{11}$

Since the balls are drawn without replacement, $5$ blue and $5$ green balls remain. The probability that the second pair consists of one blue and one green ball is:

$P_2 = \dfrac{^{5}C_{1} \times ^{5}C_{1}}{^{10}C_{2}} = \dfrac{25}{45} = \dfrac{5}{9}$

Continuing this process for all $6$ pairs, the probabilities are:

$P_3 = \dfrac{^{4}C_{1} \times ^{4}C_{1}}{^{8}C_{2}} = \dfrac{16}{28} = \dfrac{4}{7}$

$P_4 = \dfrac{^{3}C_{1} \times ^{3}C_{1}}{^{6}C_{2}} = \dfrac{9}{15} = \dfrac{3}{5}$

$P_5 = \dfrac{^{2}C_{1} \times ^{2}C_{1}}{^{4}C_{2}} = \dfrac{4}{6} = \dfrac{2}{3}$

$P_6 = \dfrac{^{1}C_{1} \times ^{1}C_{1}}{^{2}C_{2}} = \dfrac{1}{1} = 1$

The required probability is the product of these individual probabilities:

$P = P_1 \times P_2 \times P_3 \times P_4 \times P_5 \times P_6$

$P = \dfrac{6}{11} \times \dfrac{5}{9} \times \dfrac{4}{7} \times \dfrac{3}{5} \times \dfrac{2}{3} \times 1$

Canceling common factors in the numerator and denominator:

$P = \dfrac{6 \times 4 \times 2}{11 \times 9 \times 7} = \dfrac{48}{693} = \dfrac{16}{231}$

Answer: $\dfrac{16}{231}$