JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Probability

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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 :

(A) $\frac{63}{925}$
(B) $\frac{17}{231}$
(C) $\frac{16}{231}$ ✓ Correct
(D) $\frac{64}{925}$

Answer: (C)

Solution

Partition the

12

balls into

6

unordered pairs in

\frac{12 !}{2 ^{6} 6 !} = 10395

ways. A favourable partition matches each blue ball with a green ball, which is a one-to-one pairing of the

6

blue with the

6

green balls, giving

6! = 720

partitions. The probability is

\frac{720}{10395} = \frac{16}{231}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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