Probability of Drawing Two Red Balls from a Jar - A Comprehensive Guide

When dealing with probability problems, especially those involving the drawing of balls from a jar, a clear understanding of combinatorics and the concept of drawing without replacement is essential. In this article, we will explore a detailed step-by-step method to solve a specific problem, illustrating the calculation behind the probability of drawing two red balls from a jar containing a mix of red and blue balls. We will also discuss the broader implications of this problem, including related scenarios and their solutions.

Problem Overview

Consider a jar containing 4 red balls and 7 blue balls. If two balls are drawn at random without replacement, what is the probability that exactly one of the balls drawn is red?

Step-by-Step Solution

To solve this problem, we will use the basic principles of combinatorics and probability. The key steps are:

Calculate the total number of ways to draw 2 balls out of 12 without replacement. Determine the number of favorable outcomes, i.e., drawing exactly one red ball and one blue ball. Divide the number of favorable outcomes by the total number of possible outcomes to find the probability.

Step 1: Total Number of Ways to Draw 2 Balls without Replacement

The total number of ways to draw 2 balls out of 12 is given by the combination formula:

(^{12}C_2 frac{12!}{2!(12-2)!} frac{12!}{2!10!} frac{12 times 11 times 10!}{2 times 1 times 10!} 66)

Step 2: Number of Favorable Outcomes - Drawing Exactly One Red Ball and One Blue Ball

To find the number of favorable outcomes, we need to calculate the number of ways to choose 1 red ball out of 4 red balls and 1 blue ball out of 7 blue balls:

(^{4}C_1 times ^{7}C_1 frac{4!}{1!3!} times frac{7!}{1!6!} 4 times 7 28)

Step 3: Probability Calculation

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

Probability (frac{28}{66} frac{14}{33})

Additional Insights and Scenarios

The problem of drawing balls without replacement is a classic example in probability and statistics, often appearing in various contexts, such as quality control, games of chance, and lottery drawings.

Scenario 1: Drawing with the Different Number of Balls

Let's consider a scenario where the jar contains 7 red balls and 3 blue balls. The probability of drawing two red balls without replacement is calculated as follows:

Probability of the first ball being red: (frac{7}{10}) Probability of the second ball being red after the first red ball is drawn: (frac{6}{9} frac{2}{3}) Total probability of both balls being red: (frac{7}{10} times frac{6}{9} frac{7}{10} times frac{2}{3} frac{14}{30} frac{7}{15})

Scenario 2: Combined Random Draws

Another common scenario involves drawing two balls in succession, where the probability of both events occurring is calculated by multiplying the individual probabilities:

Probability of drawing the first ball red: (frac{8}{12} frac{2}{3}) Probability of drawing the second ball red after the first one is removed: (frac{7}{11}) Total probability of both balls being red: (frac{2}{3} times frac{7}{11} frac{14}{33})

Conclusion

Understanding and solving problems involving the drawing of balls from a jar is a fundamental aspect of probability and combinatorics. The principles and methods used here can be applied to a wide range of similar problems, enhancing both theoretical knowledge and practical problem-solving skills.