Problem 3 - Calculus for Biology and Medicine

Question

In Problems \(1-4\), determine the sample space for each random experiment. An urn contains five balls numbered \(1-5\), respectively. The random experiment consists of selecting two balls simultaneously without replacement.

Step-by-Step Solution

Answer

The sample space is \((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\).

Step 1: Understanding the Concept

The sample space of a random experiment refers to all possible outcomes. Here, the experiment involves selecting two balls from an urn containing five balls (numbered 1 to 5) without replacement. This means each selection will include a pair of distinct, differently numbered balls.

Step 2: Identifying Total Outcomes

Since we need to select two balls from five, we use combinations to find all possible pairs of selections. This is because the order does not matter, and each ball cannot be selected more than once.

Step 3: Calculate Combinations

Use the formula for combinations: \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n = 5 \) (the total number of balls) and \( k = 2 \) (the number of balls to be selected). Substitute the values: \( C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \). Thus, there are 10 possible pairs.

Step 4: Enumerate the Sample Space

List all possible pairs of selected balls: (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5). This list represents the entire sample space for the experiment.

Key Concepts

Combinations

In the context of probability and set theory, combinations refer to the selection of items from a larger set where the order does not matter. It's crucial to differentiate combinations from permutations, as the latter involves consideration of order. For our urn example with balls numbered from 1 to 5, we need to pick any two balls at a time. Since the order does not matter in our selection (i.e., choosing ball 1 and ball 2 is the same as choosing ball 2 and ball 1), we employ combinations. The mathematical formula used is \( C(n, k) = \frac{n!}{k!(n-k)!} \). Here, "\(!\)" denotes factorial, meaning the product of all positive integers up to that number. Consider that we only want to select 2 balls (\(k = 2\)) from the total of 5 (\(n = 5\)). Applying the formula, \(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\). This calculation reveals that there are 10 unique pairs that can be formed from the five balls. Hence, when dealing with combinations, remember that order does not affect the outcome.

Random Experiment

In probability theory, a random experiment is any process that leads to a set of possible outcomes. This is distinct from a deterministic process because each execution can yield different results. The unpredictability of outcomes is what makes the experiment "random." In our scenario of selecting balls from the urn, the act of picking two balls without looking leads to a random outcome. Each time two balls are picked, different numbered balls might be selected, creating different results. What differentiates random experiments is their lack of predetermined outcomes. Thus, every execution from the defined conditions - like picking balls without replacement from an urn containing different numbers - inherently involves chance. This random selection is fundamental in studying event probabilities and forming the "sample space."

Probability Theory

Probability theory is a branch of mathematics concerned with analyzing random events. The core idea is to measure the likelihood of an event occurring within a defined sample space. The sample space, as determined in our exercise, consists of all possible outcomes: all pairs of two balls selected from an urn containing five. These pairs describe our complete set of potential results based on the random experiment's conditions. Because the sample space has 10 pairs, recognizing the likelihood of each can guide predictions about the experiment's results. A probability measure assigns a number to each event, quantifying its likelihood between 0 and 1. If all outcomes are equally probable, the probability of picking any specific pair is \(\frac{1}{10}\). Probability theory helps us understand how often an event might happen, considering each trial is an independent event where past picks do not influence future outcomes.