Probability of Drawing Non-Blue Balls: A Detailed Analysis

In this article, we will delve into the probability of drawing non-blue balls from a box containing 6 blue, 5 green, and 4 red balls. We will explore the concepts of both with and without replacement, and calculate the exact probabilities for each scenario.

Understanding the Basics: Probability

The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. This article applies this fundamental concept to solve the problem at hand.

Solving without Replacement

Let's consider the case where the balls are drawn without replacement. This means that after each ball is drawn, it is not returned to the box.

We start by calculating the total number of possible outcomes. The box contains a total of 15 balls (6 blue, 5 green, and 4 red):

Quantity of marbles 6 5 4 15

The number of ways to choose 2 balls out of 15 is given by the combination formula:

( {}^{15}C_2 frac{15 times 14}{2} 105 )

We are interested in the probability that neither of the two balls drawn is blue. This means we are looking to draw either a green or a red ball.

The number of favorable outcomes is the number of ways to choose 2 balls from the 10 non-blue balls (5 green 4 red):

( {}^{10}C_2 frac{10 times 9}{2} 45 )

Hence, the probability of drawing 2 non-blue balls without replacement is:

( P(text{both non-blue}) frac{{}^{10}C_2}{{}^{15}C_2} frac{45}{105} frac{3}{7} approx 0.4286 )

This result can also be calculated as:

( P(text{both non-blue}) 1 - P(text{at least one blue}) 1 - frac{6}{15} 1 - frac{2}{5} frac{3}{5} )

Solving with Replacement

Now let's consider the case where the balls are drawn with replacement. This means that after each ball is drawn, it is returned to the box before drawing the second ball.

The probability of drawing a non-blue ball in the first draw is:

( P(text{1st non-blue}) frac{10}{15} frac{2}{3} )

Since the ball is replaced, the total number of balls remains the same for the second draw. The probability of drawing a non-blue ball in the second draw is also:

( P(text{2nd non-blue}) frac{10}{15} frac{2}{3} )

The probability of drawing 2 non-blue balls with replacement is then:

( P(text{both non-blue}) frac{2}{3} times frac{2}{3} frac{4}{9} approx 0.4444 )

Conclusion

From our analysis, we can conclude the following:

Without replacement, the probability of drawing 2 non-blue balls is approximately 0.4286 (or 42.86%). With replacement, the probability is about 0.4444 (or 44.44%).

These calculations highlight the difference in probability depending on whether the balls are drawn with or without replacement. Excel simulations confirm these theoretical results, providing a robust framework for understanding and solving similar probability problems.