Key Concepts and Formulas

To effectively solve this probability problem, we will utilize the following fundamental concepts from combinatorics and probability:

1. Probability Definition: The probability of an event $E$, denoted as $P(E)$, is the ratio of the number of outcomes favorable to $E$ to the total number of possible outcomes in the sample space, assuming all outcomes are equally likely. $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

2. Combinations: When the order of selection does not matter, we use combinations to count the number of ways to choose $r$ items from a set of $n$ distinct items. This is particularly relevant for problems involving drawing items simultaneously or where the final group composition is important, not the sequence of drawing. The formula for combinations is: $^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$ where $n$ is the total number of items available, and $r$ is the number of items to be chosen.

3. Multiplication Principle of Counting: If an event can be broken down into a sequence of independent choices, and there are $k_1$ ways to make the first choice, $k_2$ ways to make the second choice, and so on, then the total number of ways to perform the entire sequence of choices is the product $k_1 \times k_2 \times \dots$. This principle is crucial when combining selections from different categories (e.g., choosing one red AND one blue AND one green ball).

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Step-by-Step Solution

Let's break down the problem to systematically arrive at the solution.

Understanding the Problem: We are given an urn containing 9 balls in total, with the following distribution:

• Red balls (R): 3
• Blue balls (B): 4
• Green balls (G): 2 We are drawing 3 balls at random without replacement. This means that once a ball is drawn, it is not returned to the urn, so it cannot be drawn again. Our goal is to find the probability that the three balls drawn have different colours. This specific condition implies that we must draw exactly one red ball, one blue ball, and one green ball.

Step 1: Calculate the Total Number of Possible Outcomes (Sample Space)

• What we are doing: We need to determine the total number of unique ways to select any 3 balls from the 9 balls available in the urn. This value will serve as the denominator in our probability calculation, representing the size of the entire sample space.
• Why we use combinations: Since the problem asks for the probability of drawing a set of 3 balls (the order in which they are drawn doesn't change the composition of the set, e.g., Red-Blue-Green is the same set as Green-Red-Blue), combinations are the appropriate method for counting. Also, the drawing is "without replacement," which is inherently handled by the combination formula.
• Calculation:
  • Total number of balls, $n = 9$.
  • Number of balls to be drawn, $r = 3$.
  • The total number of ways to choose 3 balls from 9 is given by $^9C_3$. $^9C_3 = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!}$ Expanding the factorials and simplifying: $^9C_3 = \frac{9 \times 8 \times 7 \times 6!}{ (3 \times 2 \times 1) \times 6!}$ We cancel out $6!$ from the numerator and the denominator: $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1}$ $^9C_3 = \frac{504}{6}$ $^9C_3 = 84$ Thus, the total number of possible outcomes (the size of our sample space) is 84.

Step 2: Calculate the Number of Favorable Outcomes (Event Space)

• What we are doing: We need to determine the number of ways to choose 3 balls such that each ball has a different color. This means we must select exactly one red ball, one blue ball, and one green ball. This value will be the numerator in our probability calculation.
• Why we use combinations and the Multiplication Principle: To select one ball of each color, we make three independent selections: one from the red balls, one from the blue balls, and one from the green balls. We use combinations for each individual selection (e.g., choosing 1 red ball from 3 red balls) because the order of choosing within a color group doesn't matter. Then, we multiply these individual counts together using the Multiplication Principle of Counting, as these selections are independent and all must occur to satisfy the condition.
• Calculation:
  • Number of ways to choose 1 red ball from the 3 available red balls: $^3C_1 = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = \frac{3 \times 2 \times 1}{(1) \times (2 \times 1)} = 3$ (A useful shortcut: $^nC_1 = n$)
  • Number of ways to choose 1 blue ball from the 4 available blue balls: $^4C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = \frac{4 \times 3 \times 2 \times 1}{(1) \times (3 \times 2 \times 1)} = 4$
  • Number of ways to choose 1 green ball from the 2 available green balls: $^2C_1 = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = \frac{2 \times 1}{(1) \times (1)} = 2$
  • Total number of favorable outcomes = (ways to choose 1 Red) $\times$ (ways to choose 1 Blue) $\times$ (ways to choose 1 Green) $= ^3C_1 \times ^4C_1 \times ^2C_1 = 3 \times 4 \times 2 = 24$ So, the number of favorable outcomes is 24.

Step 3: Calculate the Probability

• What we are doing: We combine the results from Step 1 (total possible outcomes) and Step 2 (favorable outcomes) using the fundamental definition of probability.
• Why this is the final step: This ratio directly gives us the likelihood of the desired event occurring.
• Calculation: $P(\text{three balls have different colours}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$ $P(\text{three balls have different colours}) = \frac{24}{84}$ To simplify the fraction, we find the greatest common divisor of 24 and 84, which is 12. $P(\text{three balls have different colours}) = \frac{24 \div 12}{84 \div 12} = \frac{2}{7}$

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Common Mistakes & Tips

• Combinations vs. Permutations: A common error is confusing when to use combinations versus permutations. Remember, if the order of selection does not matter (e.g., drawing a hand of cards, selecting a group of balls), use combinations. If the order does matter (e.g., arranging letters, assigning distinct roles), use permutations. For this problem, the order of drawing the three balls is irrelevant to their final composition of colors, so combinations are correct.
• "Without Replacement": This phrase is critical. Combinations inherently account for "without replacement" because once an item is selected, it's no longer available for further selection within that specific combination. If the problem stated "with replacement," the counting method would involve powers (e.g., $n^r$) and be significantly different.
• Interpreting "Different Colours": In this context, "different colours" means one of each available color type (Red, Blue, Green). If the urn had only two colors, "different colours" would mean one of each of those two colors, and the third ball would have to match one of the first two. Always clarify the exact meaning of the event.
• Arithmetic Accuracy: Factorial calculations and simplifying fractions can be prone to errors. Always double-check your arithmetic, especially when simplifying fractions to their lowest terms.

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Summary

This problem is a classic application of probability involving combinations. The systematic approach involved first determining the total number of ways to draw 3 balls from 9 using combinations, which formed our sample space. Next, we calculated the number of ways to draw exactly one ball of each color (one red, one blue, one green) by multiplying the combinations for each individual color selection, forming our event space. Finally, we divided the number of favorable outcomes by the total number of possible outcomes to find the probability and simplified the resulting fraction.

The final answer is $\boxed{{2 \over 7}}$, which corresponds to option (A).