Probability of Drawing One Red and One Green Ball Without Replacement

Introduction

In probability theory, combinatorics is a powerful tool for calculating the likelihood of various outcomes. This article will explore how to find the probability of drawing exactly one red ball and one green ball from a set of 9 red and 12 green balls without replacement. We will walk through the steps and calculations using combinatorial methods.

Calculation Steps

To find the probability of drawing exactly one red ball and one green ball, we first calculate the total number of ways to draw 2 balls from 21 balls, then calculate the number of favorable outcomes, and finally find the probability.

Total Number of Ways to Draw 2 Balls

The total number of ways to choose 2 balls from 21 can be calculated using the combination formula:

[ binom{n}{r} frac{n!}{r!(n-r)!} ]

Here, (n 21) and (r 2). Therefore:

[ binom{21}{2} frac{21!}{2!(21-2)!} frac{21 times 20}{2 times 1} 210 ]

This means there are 210 ways to choose 2 balls from 21.

Number of Ways to Draw 1 Red and 1 Green Ball

To draw 1 red ball from 9 red balls and 1 green ball from 12 green balls, we use the combination formula again:

[ binom{9}{1} 9 ]

and

[ binom{12}{1} 12 ]

The total number of ways to draw 1 red and 1 green ball is:

[ 9 times 12 108 ]

Probability of Drawing Exactly One Red and One Green Ball

The probability of drawing exactly one red and one green ball is the ratio of the number of favorable outcomes to the total number of outcomes:

[ P(1 text{ red and } 1 text{ green}) frac{108}{210} ]

Simplifying the fraction:

[ frac{108}{210} frac{54}{105} frac{18}{35} ]

Therefore, the probability of drawing exactly one red ball and one green ball is:

[ boxed{frac{18}{35}} ]

Alternative Calculation Method

We can also calculate this by considering two scenarios:

Drawing a red ball first and then a green ball. Drawing a green ball first and then a red ball.

The probability of drawing a red ball first and then a green ball is:

[ frac{9}{21} times frac{12}{20} frac{9}{21} times frac{3}{5} frac{27}{105} ]

The probability of drawing a green ball first and then a red ball is:

[ frac{12}{21} times frac{9}{20} frac{12}{21} times frac{9}{20} frac{108}{420} ]

The total probability is the sum of these two scenarios:

[ frac{27}{105} frac{108}{420} frac{72}{420} frac{108}{420} frac{180}{420} frac{36}{84} frac{6}{14} frac{3}{7} ]

Another Calculation Method

We can calculate the probability using combinations:

We can select one red ball in (9 text{C}1) ways. We can select one green ball in (12 text{C}1) ways. We can select two balls out of 21 in (21 text{C}2) ways.

Therefore:

[ text{Required probability} frac{9text{C}1 times 12text{C}1}{21text{C}2} ]

Since (9text{C}1 9) and (12text{C}1 12), and (21text{C}2 frac{21 times 20}{2 times 1} 210), we get:

[ 9 times 12 times 2 / 21 times 20 0.51428571 ]

Thus, the probability is approximately 51.4%

Conclusion

Using combinatorics, we can find the probability of drawing one red and one green ball as (frac{18}{35}) or approximately 51.4%. This method is effective in solving similar probabilistic problems involving combinatorial outcomes.

By understanding and applying the principles of combinatorics, we can accurately calculate the probability of drawing specific combinations of balls or other objects. This skill is valuable in various fields, including mathematics, statistics, and data science.