Question

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 ball and one green ball is:

• A. \( \frac{63}{925} \)
• B. \( \frac{17}{231} \)
• C. \( \frac{16}{231} \)
• D. \( \frac{64}{925} \)

Answer 1

The Correct Option is D

Concept: The probability that each pair contains one blue and one green ball can be found by comparing:

• Total ways to divide the 12 balls into 6 unordered pairs
• Favorable ways where each pair has exactly one blue and one green ball

Step 1: {Find total ways to form 6 pairs from 12 balls.} The number of ways to divide \(12\) distinct objects into \(6\) unordered pairs is: \[ \frac{12!}{2^6\,6!} \] Step 2: {Find favourable outcomes.} Each pair must contain exactly one blue and one green ball. For each blue ball we choose one green ball. Number of ways to match \(6\) blue balls with \(6\) green balls: \[ 6! \] Step 3: {Compute the probability.} \[ P=\frac{6!}{\frac{12!}{2^6\,6!}} \] \[ =\frac{6!\times2^6\times6!}{12!} \] \[ =\frac{720\times64\times720}{479001600} \] \[ =\frac{64}{925} \]