Calculating the Probability of Drawing Two Red Balls Without Replacement: A Comprehensive Guide

Probability is a fundamental concept in statistics that helps us understand the likelihood of events occurring. In this article, we will explore a classic scenario involving a bag of balls and apply the principles of probability to solve this problem. We will calculate the probability of drawing two red balls from a bag containing a mix of red, blue, and green balls without replacement. By the end of this guide, you will have a clear understanding of how to solve similar problems and appreciate the elegance of probability theory.

Understanding the Scenario

Imagine a bag that contains 5 red balls and 3 blue balls. We will now draw two balls from the bag without replacement. This means that after drawing the first ball, it will not be placed back into the bag before drawing the second ball. We will explore the probability of drawing two red balls in this situation.

Evaluating the Probability Step by Step

Step 1: Determine the Total Number of Balls

To begin our calculation, we need to first determine the total number of balls in the bag. In this case, there are 5 red balls and 3 blue balls, giving us a total of:

5 3 8

Step 2: Calculate the Total Number of Ways to Choose 2 Balls from 8

The number of ways to choose 2 balls from 8 can be calculated using combinations. A combination is a selection of items where the order does not matter. The formula for combinations is given by:

binom{n}{r} dfrac{n!}{r!(n-r)!}

Plugging in our values, we have:

binom{8}{2} dfrac{8!}{2!(8-2)!} dfrac{8 times 7}{2 times 1} 28

Step 3: Calculate the Number of Ways to Choose 2 Red Balls from 5

Next, we need to determine the number of ways to choose 2 red balls from the 5 available red balls. Using the combination formula again:

binom{5}{2} dfrac{5!}{2!(5-2)!} dfrac{5 times 4}{2 times 1} 10

Step 4: Calculate the Probability

The probability of drawing two red balls without replacement is the ratio of the favorable outcomes (choosing 2 red balls) to the total number of outcomes (choosing 2 balls from 8). Therefore, we have:

P(both red) dfrac{text{Number of ways to choose 2 red balls}}{text{Total ways to choose 2 balls}} dfrac{10}{28} dfrac{5}{14}

Exploring Additional Scenarios

Let's consider the scenarios that occur when drawing without replacement and how they affect the probability:

After the first blue ball is drawn: There are 5 red and 2 blue balls left, making a total of 7 balls. The probability of drawing a red ball next is dfrac{5}{7}. After the first red ball is drawn: There are 4 red and 2 blue balls left, making a total of 6 balls. The probability of drawing another red ball next is dfrac{4}{6} dfrac{2}{3}. Combining the probabilities: The probability of drawing a blue ball first and then a red one is dfrac{3}{10} times dfrac{5}{9} dfrac{1}{6}. Similarly, the probability of drawing a green ball first and then a red one is dfrac{2}{10} times dfrac{5}{9} dfrac{1}{9}. The probability of drawing a blue ball first and then another red one is dfrac{3}{10} times dfrac{2}{9} dfrac{1}{15}.

Conclusion

Understanding the concept of probability and familiarizing ourselves with the rules of combinations are crucial skills in solving such problems. The probability of drawing two red balls from the bag without replacement is dfrac{5}{14}. This detailed guide provides a step-by-step approach to solving similar problems, allowing readers to apply the same techniques to other scenarios involving probability and combinations.

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