JEE Main 2026 — Probability Question with Solution

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

$P_1=\frac{{}^6{C}_1{\times }^6{C}_1}{{}^{12}{C}_2}=\frac{36}{66}=\frac{6}{11}$

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:

$P_2=\frac{{}^5{C}_1{\times }^5{C}_1}{{}^{10}{C}_2}=\frac{25}{45}=\frac{5}{9}$

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

$P_3=\frac{{}^4{C}_1{\times }^4{C}_1}{{}^8{C}_2}=\frac{16}{28}=\frac{4}{7}$

$P_4=\frac{{}^3{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{9}{15}=\frac{3}{5}$

$P_5=\frac{{}^2{C}_1{\times }^2{C}_1}{{}^4{C}_2}=\frac{4}{6}=\frac{2}{3}$

$P_6=\frac{{}^1{C}_1{\times }^1{C}_1}{{}^2{C}_2}=\frac{1}{1}=1$

The required probability is the product of these individual probabilities:

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

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

Canceling common factors in the numerator and denominator:

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

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