Q.9. A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the

Q: What is the correct answer for JEE Main Mathematics Q9?

The correct answer is C. Sol. Total number of balls = 12 (6 blue, 6 green). We draw pairs of balls without replacement until the bag is empty. This means 6 pairs are drawn. We want the probability that each drawn pair consists of one blue and one green ball. This is equivalent to arranging 6 B's and 6 G's such that B and G alternate. Consider the sequence of 12 draws. The first ball can be B or G. If it's B, the second mu

Q: Q.9. 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 :

The correct answer is C. Sol. Total number of balls = 12 (6 blue, 6 green). We draw pairs of balls without replacement until the bag is empty. This means 6 pairs are drawn. We want the probability that each drawn pair consists of one blue and one green ball. This is equivalent to arranging 6 B's and 6 G's such that B and G alternate. Consider the sequence of 12 draws. The first ball can be B or G. If it's B, the second mu

Q: What are the options for this JEE Main Mathematics question?

A. $\frac{63}{925}$ · B. $\frac{17}{231}$ · C. $\frac{16}{231}$ · D. $\frac{64}{925}$

Question

Q.9. 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 :

FirstInTest · Accepted Answer

Sol. Total number of balls = 12 (6 blue, 6 green). We draw pairs of balls without replacement until the bag is empty. This means 6 pairs are drawn. We want the probability that each drawn pair consists of one blue and one green ball. This is equivalent to arranging 6 B's and 6 G's such that B and G alternate. Consider the sequence of 12 draws. The first ball can be B or G. If it's B, the second must be G. If it's G, the second must be B. This means the sequence must be BGBGBGBGBGBG or GBGBGBGBGBGB. The total number of ways to arrange 6 B's and 6 G's is $\binom{12}{6}$. The number of ways to have alternating colors is 2 (BGBG... or GBGB...). However, this is not about arrangement, but about drawing pairs. The problem is equivalent to finding the probability that when we draw 6 pairs, each pair is (B, G). Let's consider the first pair. The probability of drawing (B, G) in any order is $P(BG 	ext{ or } GB) = P(B_1G_2) + P(G_1B_2) = \frac{6}{12} \cdot \frac{6}{11} + \frac{6}{12} \cdot \frac{6}{11} = 2 \cdot \frac{36}{132} = \frac{72}{132} = \frac{6}{11}$. After drawing one (B, G) pair, we have 5 B and 5 G balls left (10 balls). The probability of drawing the second (B, G) pair is $\frac{5}{10} \cdot \frac{5}{9} + \frac{5}{10} \cdot \frac{5}{9} = 2 \cdot \frac{25}{90} = \frac{5}{9}$. This continues until the last pair. The probability is the product of these probabilities: $P = \frac{6}{11} \cdot \frac{5}{9} \cdot \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1}$. This is incorrect. This is for drawing specific pairs. The problem is about drawing 6 pairs, and each pair is (B,G). This is a classic problem. The total number of ways to draw 6 pairs from 12 balls is $\frac{\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{6!}$. The number of ways to draw 6 pairs, each consisting of one blue and one green ball, is $\binom{6}{1}\binom{6}{1} 	imes \binom{5}{1}\binom{5}{1} 	imes \dots 	imes \binom{1}{1}\binom{1}{1} = (6!)^2$. The total number of ways to draw 6 pairs is $\frac{\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{6!} = \frac{66 	imes 45 	imes 28 	imes 15 	imes 6 	imes 1}{720} = 10395$. This is not the standard approach. Let's use the solution's approach. The solution states the probability is $\frac{6 	imes 6}{12 	imes 11} 	imes \frac{5 	imes 5}{10 	imes 9} 	imes \frac{4 	imes 4}{8 	imes 7} 	imes \frac{3 	imes 3}{6 	imes 5} 	imes \frac{2 	imes 2}{4 	imes 3} 	imes \frac{1 	imes 1}{2 	imes 1}$. This is the probability of drawing a BG pair, then another BG pair, and so on, in that specific order. This is for the case where the first ball is B and the second is G for each pair. If the order within the pair doesn't matter, then the probability of drawing one B and one G in the first draw is $\frac{\binom{6}{1}\binom{6}{1}}{\binom{12}{2}} = \frac{36}{66} = \frac{6}{11}$. For the second draw, $\frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{25}{45} = \frac{5}{9}$. For the third draw, $\frac{\binom{4}{1}\binom{4}{1}}{\binom{8}{2}} = \frac{16}{28} = \frac{4}{7}$. For the fourth draw, $\frac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}} = \frac{9}{15} = \frac{3}{5}$. For the fifth draw, $\frac{\binom{2}{1}\binom{2}{1}}{\binom{4}{2}} = \frac{4}{6} = \frac{2}{3}$. For the sixth draw, $\frac{\binom{1}{1}\binom{1}{1}}{\binom{2}{2}} = \frac{1}{1} = 1$. The product of these probabilities is $\frac{6}{11} 	imes \frac{5}{9} 	imes \frac{4}{7} 	imes \frac{3}{5} 	imes \frac{2}{3} 	imes 1 = \frac{6 	imes 5 	imes 4 	imes 3 	imes 2}{11 	imes 9 	imes 7 	imes 5 	imes 3} = \frac{720}{10395} = \frac{16}{231}$. This matches option (3). The solution in the image has $\frac{6 	imes 6}{12 	imes 11}$ etc. This is for drawing one B then one G, or one G then one B. The probability of drawing a B then a G is $\frac{6}{12} 	imes \frac{6}{11}$. The probability of drawing a G then a B is $\frac{6}{12} 	imes \frac{6}{11}$. So the probability of drawing one B and one G in a specific order (e.g., B then G) is $\frac{36}{132}$. The probability of drawing one B and one G (order doesn't matter) is $\frac{\binom{6}{1}\binom{6}{1}}{\binom{12}{2}} = \frac{36}{66} = \frac{6}{11}$. The solution's calculation is $\frac{6 	imes 6 	imes 5 	imes 5 	imes 4 	imes 4 	imes 3 	imes 3 	imes 2 	imes 2 	imes 1 	imes 1}{12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = \frac{(6!)^2}{(12)!} 	imes 2^6$. This is not quite right. The solution's calculation is $\frac{6 	imes 6}{12 	imes 11} 	imes \frac{5 	imes 5}{10 	imes 9} 	imes \frac{4 	imes 4}{8 	imes 7} 	imes \frac{3 	imes 3}{6 	imes 5} 	imes \frac{2 	imes 2}{4 	imes 3} 	imes \frac{1 	imes 1}{2 	imes 1} = \frac{36}{132} 	imes \frac{25}{90} 	imes \frac{16}{56} 	imes \frac{9}{30} 	imes \frac{4}{12} 	imes \frac{1}{2} = \frac{3}{11} 	imes \frac{5}{18} 	imes \frac{2}{7} 	imes \frac{3}{10} 	imes \frac{1}{3} 	imes \frac{1}{2} = \frac{16}{231}$. This is the probability of drawing a blue ball first, then a green ball, for each of the 6 pairs. If the question implies that each pair *consists* of one blue and one green, without specifying order, then my calculation of $\frac{16}{231}$ is correct. The solution's calculation also yields $\frac{16}{231}$.

Answer

$\frac{63}{925}$

Answer

$\frac{17}{231}$

Answer

$\frac{16}{231}$

Answer

$\frac{64}{925}$

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