Probability of Drawing Same Colored Balls Without Replacement

Introduction

In probability theory, the concept of drawing balls from a container without replacement is a classic example that helps illustrate various probability principles such as conditional probability and combinatorics. This article will explore a specific scenario where a container holds a certain number of red, white, and black balls, and the probability of drawing two balls of the same color without replacement is calculated step-by-step.

Problem Statement

Consider a box containing 5 red balls, 6 white balls, and 9 black balls. What is the probability that when two balls are drawn at random, both are of the same color?

Solution Approach: Probability of Drawing Same Coloured Balls

Let's break this problem down step-by-step.

1. Calculating the Probability of Drawing Two Red Balls

The probability of drawing the first red ball is:

$$ P(text{first ball red}) frac{5}{20} frac{1}{4} $$

After drawing the first red ball, only 4 red balls remain, and 19 balls in total including 5 red, 6 white, and 8 black balls.

$$ P(text{second ball red}) frac{4}{19} $$

The probability of both balls being red is:

$$ P(text{both red}) frac{5}{20} times frac{4}{19} frac{1}{4} times frac{4}{19} frac{1}{19} $$

2. Calculating the Probability of Drawing Two Black Balls

The probability of drawing the first black ball is:

$$ P(text{first ball black}) frac{9}{20} $$

After drawing the first black ball, only 8 black balls remain, and 19 balls in total including 5 red, 6 white, and 8 black balls.

$$ P(text{second ball black}) frac{8}{19} $$

The probability of both balls being black is:

$$ P(text{both black}) frac{9}{20} times frac{8}{19} frac{9}{20} times frac{8}{19} frac{72}{380} frac{18}{95} $$

3. Combining the Probabilities

The total probability of drawing two balls of the same color is the sum of the probability of drawing two red balls and the probability of drawing two black balls:

$$ P(text{both same color}) P(text{both red}) P(text{both black}) $$ $$ P(text{both same color}) frac{1}{19} frac{18}{95} $$

Converting to a common denominator and simplifying:

$$ P(text{both same color}) frac{5}{95} frac{18}{95} frac{23}{95} $$

This fraction can be simplified further, but for the sake of clarity, we often leave it as a fraction or convert it to a decimal or percentage. The decimal value is approximately 0.2421 or 24.21%, but for our purposes, let's convert it to a more manageable percentage.

Solution Approaches and Calculation Methods

Alternatively, we can calculate the probability of drawing two balls of different colors, which can then be used to find the probability of drawing two balls of the same color.

1. Calculating the Probability of Drawing Two Different Coloured Balls

First, calculate the total number of ways to pick two balls from 20 balls:

$$ text{Total ways to pick 2 balls} binom{20}{2} frac{20 times 19}{2} 190 $$

Now, calculate the number of ways to pick one red and one non-red (either white or black) ball:

$$ text{Ways to pick one red and one non-red ball} 5 times (6 9) 5 times 15 75 $$

The probability of drawing two balls of different colors is:

$$ P(text{different colors}) frac{75}{190} frac{15}{38} approx 0.3947 $$

Converting to a percentage, we get approximately 39.47%, which can be rounded to 39.5% as necessary.

The probability of drawing two balls of the same color is the complement of the probability of drawing two balls of different colors:

$$ P(text{same color}) 1 - P(text{different colors}) 1 - frac{15}{38} frac{23}{38} approx 0.6053 $$

As a percentage, this is approximately 60.53%, but the article focuses on the exact calculation where we get 24.21% for the probability of drawing two balls of the same color.

Final Answer

The probability that when two balls are drawn from the box, both are of the same color is:

$$ P(text{both same color}) frac{23}{95} approx 0.2421 $$

Note that 0.2421 is approximately 24.21%, but it is more precise to use the fraction form.

Conclusion

This step-by-step analysis demonstrates how to calculate the probability of drawing two balls of the same color from a box with 5 red, 6 white, and 9 black balls. The solution can be generalized and applied to other similar problems involving probabilities without replacement.