Understanding Probability: Drawing Two Balls from a Bag

This article discusses the problem of drawing two balls from a bag containing 3 red balls, 3 green balls, and 4 black balls. We will explore several scenarios and calculate the probabilities involved. This content provides a comprehensive analysis of the drawing problem, offering a clear and detailed explanation suitable for SEO purposes.

Introduction to the Problem

We have a bag containing three types of balls: 3 red, 3 green, and 4 black. The task is to determine the probability of various outcomes when two balls are drawn from the bag. This article will cover several specific scenarios, including drawing two red balls, drawing one green and one black ball, and drawing two non-red balls. The detailed solution approach will involve combinatorial mathematics using binomial coefficients.

Calculating the Probability of Drawing Two Red Balls

To find the probability that both balls drawn are red, we can follow these steps:

Calculate the total number of balls in the bag:

Total 3 (red) 3 (green) 4 (black) 10

Find the total number of ways to choose 2 balls from 10:

Total ways C(10, 2) (10 × 9) / (2 × 1) 45

Calculate the number of ways to choose 2 red balls from 3:

Ways to choose 2 red balls C(3, 2) (3 × 2) / (2 × 1) 3

Compute the probability:

P(both red) (Ways to choose 2 red) / (Total ways to choose 2 balls) 3 / 45 1 / 15

The probability that both balls drawn are red is 1 / 15.

Without Replacement: Drawing Two Red Balls

Let's consider the scenario where we first draw a red ball, then draw a red ball without replacement. We will break this down step-by-step:

The probability of drawing a red ball in the first draw is:

3 / 10

The probability of drawing a red ball in the second draw, given the first was red, is:

2 / 9

The total probability of drawing two red balls in sequence is the product of these probabilities:

(3 / 10) × (2 / 9) 6 / 90 1 / 15

Thus, the probability that both balls drawn are red without replacement is 1 / 15.

Probability of Drawing One Green and One Black Ball

We need to consider the different orders: first green, then black; or first black, then green:

Probability of first drawing a green ball and then a black ball:

(3 / 10) × (4 / 9) 12 / 90

Probability of first drawing a black ball and then a green ball:

(4 / 10) × (3 / 9) 12 / 90

Total requested probability, accounting for both sequences:

12 / 90 12 / 90 24 / 90 4 / 15 0.267 (approximately)

General Probability Calculations

Using combinatorial methods, we can consider more generalized scenarios. For instance, the chance to draw the two red balls from the total 10 balls is calculated as follows:

There are C(5, 2) 10 ways to choose any 2 red balls from 5. There are C(7, 1) 7 ways to choose one non-red ball from the remaining 7. Thus, there are 10 × 7 70 ways to draw 2 red and 1 non-red ball. Considering all possible ways to draw 3 balls out of 10, there are C(12, 3) 220 ways. The probability of the favorable event is:

70 / 220 7 / 22 0.318 (approximately)

In conclusion, the probability for each scenario varies, but understanding the combinatorial approach provides a solid foundation for solving these types of problems.