JEE Main 2026 — Mathematics Probability & Statistics

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

$$P_1 = \dfrac{^{6}C_{1} \times ^{6}C_{1}}{^{12}C_{2}} = \dfrac{36}{66} = \dfrac{6}{11}$$P1​=12C2​6C1​×6C1​​=6636​=116​

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

$$P_2 = \dfrac{^{5}C_{1} \times ^{5}C_{1}}{^{10}C_{2}} = \dfrac{25}{45} = \dfrac{5}{9}$$P2​=10C2​5C1​×5C1​​=4525​=95​

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

$$P_3 = \dfrac{^{4}C_{1} \times ^{4}C_{1}}{^{8}C_{2}} = \dfrac{16}{28} = \dfrac{4}{7}$$P3​=8C2​4C1​×4C1​​=2816​=74​

$$P_4 = \dfrac{^{3}C_{1} \times ^{3}C_{1}}{^{6}C_{2}} = \dfrac{9}{15} = \dfrac{3}{5}$$P4​=6C2​3C1​×3C1​​=159​=53​

$$P_5 = \dfrac{^{2}C_{1} \times ^{2}C_{1}}{^{4}C_{2}} = \dfrac{4}{6} = \dfrac{2}{3}$$P5​=4C2​2C1​×2C1​​=64​=32​

$$P_6 = \dfrac{^{1}C_{1} \times ^{1}C_{1}}{^{2}C_{2}} = \dfrac{1}{1} = 1$$P6​=2C2​1C1​×1C1​​=11​=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=P1​×P2​×P3​×P4​×P5​×P6​

$$P = \dfrac{6}{11} \times \dfrac{5}{9} \times \dfrac{4}{7} \times \dfrac{3}{5} \times \dfrac{2}{3} \times 1$$P=116​×95​×74​×53​×32​×1

Canceling common factors in the numerator and denominator:

$$P = \dfrac{6 \times 4 \times 2}{11 \times 9 \times 7} = \dfrac{48}{693} = \dfrac{16}{231}$$P=11×9×76×4×2​=69348​=23116​

Answer: $\dfrac{16}{231}$23116​