JEE Main 2026 Apr 6 Shift 2, Mathematics Q9: Introduction To Probability

JEE Main2026Apr 6, Shift 2Mathematics

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
$\frac{16}{231}$
D
$\frac{64}{925}$

Solution

Total ways of partitioning 12 balls into 6 unordered pairs: $\frac{12 !}{( 2 ! ) ^{6} 6 !}$ .

Favourable: pair each of the 6 blue with a green. Match 6 blue with 6 green: $6!$ ways.

$P = \frac{6 !}{\frac{12 !}{( 2 ! ) ^{6} 6 !}} = \frac{6 ! \cdot 6 ! \cdot 2 ^{6}}{12 !}$ .

Compute: $\frac{720 \cdot 720 \cdot 64}{479001600} = \frac{16}{231}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q9: Introduction To Probability

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