Maths Q9 – JEE-MAIN 2026

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 :

Clue 1 of 3: Sequential Probability

Consider the probability of drawing a specific type of pair at each step, given the remaining balls.

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Step 1: Define the event and initial conditions✦ Active

We have 6 blue (B) and 6 green (G) balls, totaling 12 balls. We need to draw 6 pairs, with each pair consisting of one blue and one green ball. This is a sequence of 6 draws of pairs, where the composition of the bag changes after each draw.

Step 2: Calculate the probability for each successive pair○ Expand

The probability of drawing one blue and one green ball for the first pair ( $P_{1}$ ) from 12 balls (6B, 6G) is:

P_{1} = \frac{(\binom{6}{1}) (\binom{6}{1})}{(\binom{12}{2})} = \frac{6 \times 6}{\frac{12 \times 11}{2}} = \frac{36}{66} = \frac{6}{11}

For the second pair ( $P_{2}$ ), there are 5 blue and 5 green balls left (10 total):

P_{2} = \frac{(\binom{5}{1}) (\binom{5}{1})}{(\binom{10}{2})} = \frac{5 \times 5}{\frac{10 \times 9}{2}} = \frac{25}{45} = \frac{5}{9}

Similarly, for the subsequent pairs:

P_{3} = \frac{(\binom{4}{1}) (\binom{4}{1})}{(\binom{8}{2})} = \frac{4 \times 4}{\frac{8 \times 7}{2}} = \frac{16}{28} = \frac{4}{7}

P_{4} = \frac{(\binom{3}{1}) (\binom{3}{1})}{(\binom{6}{2})} = \frac{3 \times 3}{\frac{6 \times 5}{2}} = \frac{9}{15} = \frac{3}{5}

P_{5} = \frac{(\binom{2}{1}) (\binom{2}{1})}{(\binom{4}{2})} = \frac{2 \times 2}{\frac{4 \times 3}{2}} = \frac{4}{6} = \frac{2}{3}

For the sixth pair ( $P_{6}$ ), there is 1 blue and 1 green ball left (2 total):

P_{6} = \frac{(\binom{1}{1}) (\binom{1}{1})}{(\binom{2}{2})} = \frac{1 \times 1}{1} = 1

Step 3: Calculate the total probability○ Expand

The total 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

Cancel common terms in the numerator and denominator:

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

P = \frac{48}{693}

Divide both numerator and denominator by their greatest common divisor, which is 3:

P = \frac{48 \div 3}{693 \div 3} = \frac{16}{231}

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Sequential Probability

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