A bag contains 8 red balls and 4 green balls. Three balls are drawn at random from the bag without replacement. What is the probability that exactly two of the drawn balls are red and the remaining one ball is green?

Introduction / Context:
This question focuses on probability with combinations and selections without replacement. Instead of the order of drawing, we care only about the composition of the drawn set: exactly two red balls and one green ball. Such problems test your ability to count combinations and to form a correct probability ratio based on favorable and total outcomes.

Given Data / Assumptions:

• Number of red balls in the bag = 8
• Number of green balls in the bag = 4
• Total balls in the bag = 12
• Three balls are drawn at random without replacement.
• We want exactly 2 red balls and 1 green ball among the three drawn.

Concept / Approach:
When order does not matter, combinations are used. The probability is computed as: Probability = (Number of favorable combinations) / (Total number of possible combinations). Favorable combinations correspond to choosing 2 red balls from 8 and 1 green ball from 4. The total combinations correspond to choosing any 3 balls from 12, irrespective of color. This is a direct application of the combination formula nCr.

Step-by-Step Solution:
Step 1: Compute total ways to choose any 3 balls from 12 using 12C3. 12C3 = 12 * 11 * 10 / (3 * 2 * 1) = 220. Step 2: Compute ways to choose 2 red balls from 8 using 8C2. 8C2 = 8 * 7 / (2 * 1) = 28. Step 3: Compute ways to choose 1 green ball from 4 using 4C1. 4C1 = 4. Step 4: Multiply to get total favorable combinations. Favorable combinations = 8C2 * 4C1 = 28 * 4 = 112. Step 5: Form the probability ratio using combinations. Probability = Favorable / Total = 112 / 220. Step 6: Simplify the fraction 112 / 220 by dividing numerator and denominator by 4 to get 28 / 55.

Verification / Alternative check:
You can verify using stepwise conditional probability. For example, choose an order such as red, red, green and compute (8/12) * (7/11) * (4/10). Then multiply by the number of distinct permutations of the pattern R R G, which is 3 because the green ball can appear in any of the three positions. That gives: (8/12 * 7/11 * 4/10) * 3 = (8 * 7 * 4 * 3) / (12 * 11 * 10) = 672 / 1320 which simplifies to 28 / 55, confirming the previous result.

Why Other Options Are Wrong:
Option 21/55 is smaller and comes from incorrect counting or missing some favorable arrangements. Option 32/55 overestimates the probability and could result from including outcomes with three red balls or other incorrect combinations. Option 14/55 is exactly half of the correct probability, which can happen if the factor from permutations is missed. Option 7/33 does not match any correct simplification from the standard methods used here.

Common Pitfalls:
A common mistake is to treat the events as ordered and forget to multiply by the number of permutations, or the reverse: to use combinations for the numerator and a power such as 12^3 for the denominator, which corresponds to ordered draws with replacement. Another frequent error is to misread the phrase exactly two red balls and one green ball and accidentally include the case of three red balls. Always check whether the problem cares about exact counts or at least counts, and whether selection is with or without replacement.

Final Answer:
The probability that exactly two drawn balls are red and one is green is 28/55.