Sample Space for Drawing Two Defective Items from a Box

Introduction

Understanding the concepts of sample space, probability, and permutations is crucial in the field of mathematics and statistics. In this article, we explore the sample space and probabilities associated with drawing two defective items from a box containing a mix of defective and non-defective items.

Concept Overview

The sample space in probability theory refers to the set of all possible outcomes of an experiment. In the scenario of drawing items from a box, the sample space represents all the different ways in which two items can be selected from the box. This concept is particularly important in understanding the likelihood of specific outcomes, such as drawing two defective items.

Experimental Setup

Consider a box containing five items: two defective items (denoted as X and Y) and three non-defective items (denoted as a, b, and c). The task is to determine the sample space when drawing any two items from the box without replacement.

The Sample Space

The sample space for this experiment can be constructed by listing all possible combinations of the two defective and non-defective items, considering the order in which they are drawn. Here is the sample space:

ab, ac, aX, aY ba, bc, bX, bY ca, cb, cX, cY Xa, Xb, Xc, XY Ya, Yb, Yc, YX

This results in a total of twenty possible outcomes, making the sample space of the experiment include twenty distinct ordered pairs. The sample space is crucial for determining the probability of any given event.

Probability Calculations

With the sample space established, we can calculate the probability of specific outcomes. For instance, the probability that both items drawn are defective is determined as follows:

Direct Method

The probability of drawing the first defective item (X) from the box is 2/5. After drawing X, the probability of drawing the second defective item (Y) is 1/4. Therefore, the probability of drawing both defective items (X and Y) in that order is:

P(X then Y) (2/5) × (1/4) 2/20 1/10

Since drawing Y then X also qualifies as drawing two defective items, the probability is:

P(Y then X) (2/5) × (1/4) 2/20 1/10

Adding these probabilities together gives:

P(both defective) P(X then Y) P(Y then X) 2/20 2/20 4/20 1/5

Indirect Method

Alternatively, we can determine the probability of drawing two defective items by counting the favorable outcomes in the sample space and dividing by the total number of outcomes:

Total number of outcomes (sample space) 20

Favorable outcomes (both defective): aX, aY, bX, bY, cX, cY, XY, YX

Number of favorable outcomes 8

Therefore, the probability of drawing two defective items is:

P(both defective) 8/20 1/5 0.2

Conclusion

The concepts of sample space and probability are fundamental to statistical analysis and decision-making in various fields. Understanding the process of constructing the sample space and calculating probabilities is essential for mastering these concepts.