An Urn Contains 5 Red, 6 Blue, And 8 Green Balls. If A Set Of 3 Balls Is Randomly Selected, What Is The comprehensive guide to understanding the probabilities, calculations, and concepts involved in this classic problem in combinatorics and probability theory. This article aims to break down the problem, explore various scenarios, and provide detailed solutions, making it an ideal resource for students, educators, and enthusiasts interested in probability calculations involving urns and random selections.
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Understanding the Problem: The Basics
Before diving into the calculations, it’s essential to understand the problem’s components and what is being asked.
Details of the Urn and Balls
- The urn contains a total of 19 balls:
- 5 Red balls
- 6 Blue balls
- 8 Green balls
- The task involves randomly selecting 3 balls from the urn without replacement (assuming standard probability rules unless specified otherwise).
What Is Being Asked?
Typically, problems like this ask for one or more of the following:
- The probability of selecting a specific combination of balls (e.g., 2 Red and 1 Blue).
- The probability of selecting at least one ball of a certain color.
- The probability of selecting all balls of different colors.
- General probability calculations related to the composition of the selected balls.
In this case, the phrase "what is the" suggests that the question might be incomplete or that it’s leading to a specific probability calculation, such as:
- What is the probability that all three balls are of different colors?
- What is the probability of selecting exactly two red balls?
- What is the probability of selecting at least one green ball?
For the sake of this comprehensive guide, we will explore various common scenarios and how to calculate their probabilities, providing a complete overview.
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Key Concepts in Probability and Combinatorics
Understanding the core principles is essential before tackling specific calculations.
Basic Probability Principles
- Sample Space (Total Outcomes): The total number of ways to select 3 balls from 19.
- Favorable Outcomes: The number of ways that satisfy the specific condition (e.g., selecting 2 red and 1 blue).
- Probability Formula:
Combinatorial Calculations
- Combination Formula:
- Used to calculate the number of ways to choose k items from n without regard to order.
Calculating Total Number of Ways to Select 3 Balls
The first step in any probability calculation is determining the total number of possible outcomes.
Total Combinations
Since the balls are distinct only by color, and the selection is without replacement, the total number of ways to choose 3 balls from 19 is:
\[
\text{Total outcomes} = \binom{19}{3} = \frac{19!}{3!(19-3)!} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969
\]
This total forms the denominator for all probability calculations that involve selecting any 3 balls.
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Scenarios and Probabilities
Below are common scenarios and how to compute their probabilities.
1. Probability of Selecting Three Balls of Different Colors
Objective: Find the probability that the three balls are each of different colors (i.e., one Red, one Blue, and one Green).
Step-by-step solution:
- Calculate the number of favorable outcomes:
- Ways to choose 1 Red from 5:
\[
\binom{5}{1} = 5
\]
- Ways to choose 1 Blue from 6:
\[
\binom{6}{1} = 6
\]
- Ways to choose 1 Green from 8:
\[
\binom{8}{1} = 8
\]
- Total favorable outcomes:
\[
\binom{5}{1} \times \binom{6}{1} \times \binom{8}{1} = 5 \times 6 \times 8 = 240
\]
- Probability:
\[
P(\text{all different colors}) = \frac{240}{969} \approx 0.2477
\]
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2. Probability of Selecting Two Red Balls and One Blue Ball
Objective: Find the probability of choosing exactly 2 Red and 1 Blue.
Step-by-step solution:
- Ways to choose 2 Red from 5:
\[
\binom{5}{2} = 10
\]
- Ways to choose 1 Blue from 6:
\[
\binom{6}{1} = 6
\]
- Total favorable outcomes:
\[
10 \times 6 = 60
\]
- Probability:
\[
P(\text{2 Red, 1 Blue}) = \frac{60}{969} \approx 0.0619
\]
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3. Probability of Selecting All Green Balls
Objective: Find the probability that all three selected balls are green.
Step-by-step solution:
- Ways to choose 3 Green from 8:
\[
\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
\]
- Probability:
\[
P(\text{all green}) = \frac{56}{969} \approx 0.0578
\]
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4. Probability of Selecting Exactly One Red and Two Green Balls
Objective: Find the probability of exactly 1 Red and 2 Green.
Step-by-step solution:
- Ways to choose 1 Red from 5:
\[
\binom{5}{1} = 5
\]
- Ways to choose 2 Green from 8:
\[
\binom{8}{2} = \frac{8 \times 7}{2} = 28
\]
- Total favorable outcomes:
\[
5 \times 28 = 140
\]
- Probability:
\[
P(\text{1 Red, 2 Green}) = \frac{140}{969} \approx 0.1445
\]
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Additional Probabilities and Considerations
Apart from the specific scenarios, other interesting probability questions include:
1. Probability of Selecting At Least One Green Ball
This can be approached using the complement rule:
- Complement: No green balls are selected (i.e., all selected balls are Red or Blue).
- Calculate the probability of selecting no green balls:
- Total balls without green: 5 Red + 6 Blue = 11
- Number of ways to choose 3 balls from these 11:
\[
\binom{11}{3} = \frac{11 \times 10 \times 9}{6} = 165
\]
- Probability of no green:
\[
P(\text{no green}) = \frac{165}{969}
\]
- Probability of at least one green:
\[
P(\text{at least one green}) = 1 - P(\text{no green}) = 1 - \frac{165}{969} = \frac{969 - 165}{969} = \frac{804}{969} \approx 0.8304
\]
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2. Expected Number of Green Balls in a Random Sample of 3
The expected value (mean) of green balls in the sample can be calculated as:
\[
E(\text{green}) = 3 \times \frac{\text{Total green balls}}{\text{Total balls}} = 3 \times \frac{8}{19} \approx 1.263
\]
This indicates that, on average, in many such samples, about 1.26 green balls would be selected.
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Summary of Key Probabilities
| Scenario | Probability |
|--------------------------------------------------------|--------------------------------------------|
| All three different colors | \(\frac{240}{969} \approx 0.2477\) |
| Two Red, one Blue | \(\frac{60}{969} \approx 0.0619\) |
| All green | \(\frac{56}{969} \approx 0.0578\) |
| One Red, two Green | \(\frac{140}{969} \approx 0.1445\) |
| At least one green | \(\frac{804}{969} \approx 0.8304\) |
| No green (all Red or Blue) | \(\frac{165}{969} \approx 0.1703\) |
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