An urn contains nine balls of which three are red, four are blue and two are green. Three balls are

Q: Why do we use combinations ($^nC_r$) instead of permutations ($^nP_r$)?

Great question! Permutations are for when the order of selection matters—like arranging people in a line. Here, we are just grabbing a handful of balls. Whether you pick Red then Blue then Green, or Green then Blue then Red, the final set of three balls is the same. Thus, order is irrelevant, and combinations are the correct tool.

Q: Why do we multiply the combinations for the favorable outcomes?

Think of the selection process as a sequence of independent decisions. You need to pick a Red ball AND a Blue ball AND a Green ball. In combinatorics, the word 'AND' translates to multiplication. We are essentially building a single favorable outcome by combining one choice from each color group.

Q: Can I solve this by calculating probabilities for each draw individually?

Yes, you can! You could calculate the probability of picking (R, B, G) in one specific order and then multiply by the number of possible permutations of those colors (3! = 6). However, using the combination method is often less prone to errors because it handles the 'order doesn't matter' aspect automatically.

Question

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

Options:

(A) 2/7

(B) 1/21

(C) 2/23

(D) 1/3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of an urn containing a hidden world of color: three red balls, four blue balls, and two green balls. You are tasked with reaching in and pulling out three balls at random.

The goal is to determine the probability that you pull out three balls, each of a different color. This process serves as a fundamental lesson in organizing the chaos of random events.

### Phase 1: The Total Sample Space

Before calculating the probability of our specific event, we must define the entire universe of possibilities. We have a total of $3 + 4 + 2 = 9$ balls.

Since we are drawing three balls without replacement and the order of selection does not matter, we utilize the combination formula:

$$^nC_r = \frac{n!}{r!(n-r)!}$$

To find the total number of ways to choose any three balls from nine, we calculate $^9C_3$:

$$^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$$

There are exactly **84** different ways to grab three balls from this urn. This value serves as our denominator.

### Phase 2: The Favorable Event

Now, we focus on the specific outcome where we select one red ball, one blue ball, and one green ball. This is an 'AND' condition requiring simultaneous selection.

We calculate the number of ways to choose one ball from each color group:

$$^3C_1 = 3, \quad ^4C_1 = 4, \quad ^2C_1 = 2$$

Because these events must occur together to form our desired set, we multiply the individual combinations:

$$3 	imes 4 	imes 2 = 24$$

There are **24** distinct ways to achieve our goal.

### Phase 3: The Final Calculation

Probability is defined as the ratio of the favorable outcomes to the total number of outcomes. We set up the ratio as follows:

$$P = \frac{24}{84}$$

To simplify, we divide both the numerator and the denominator by their greatest common divisor, which is 12:

$$P = \frac{24 \div 12}{84 \div 12} = \frac{2}{7}$$

The final probability of drawing three balls of different colors is **$\frac{2}{7}$**.

### The Takeaway

Mathematics is the art of counting without having to list every single possibility. By using combinations, we bypassed the need to enumerate every outcome manually.

We identified the total sample space, isolated the favorable space, and derived the elegant ratio that governs this event. Keep practicing this logic, and you will find that even the most complex probability problems become clear.

Animated Solution for Mathematics - Probability: An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

Select Answer:

Visualized Solution

Visualizing the Urn Setup

The Probability Formula

Total Outcomes: $^{9} C_{3}$ Setup

Computing $^{9} C_{3}$

Favorable Outcomes: One of Each Color

Setting up Favorable Ways

Computing Favorable Outcomes

Calculating the Probability

Final Answer & Key Takeaway

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Takeaway

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