Problem 35 - Algebra for College Students

Question

From a survey of 1000 college students, it was found that 450 of them owned cars, 700 of them owned sound systems, and 200 of them owned both a car and a sound system. If a student is chosen at random from the 1000 students, find the probability that the student (a) owns a car, given the fact that he or she owns a sound system, and (b) owns a sound system, given the fact that he or she owns a car. (a) \(\frac{2}{7}\) (b) \(\frac{4}{9}\)

Step-by-Step Solution

Answer

(a) \( \frac{2}{7} \), (b) \( \frac{4}{9} \).

Step 1: Identify Given and Required Information

We need to find two probabilities based on the given survey data: (a) the probability of owning a car given the student owns a sound system, and (b) the probability of owning a sound system given the student owns a car. The given numbers are: 450 own cars, 700 own sound systems, 200 own both.

Step 2: Calculate Probability for Part (a)

For part (a), we need to find the probability that a student owns a car given that they own a sound system. This is a conditional probability, expressed as \( P(A|B) \), where \( A \) is owning a car and \( B \) is owning a sound system. The formula is:\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]Where \( P(A \cap B) \) is the probability of owning both a car and a sound system, and \( P(B) \) is the probability of owning a sound system. We calculate:\[ P(A \cap B) = \frac{200}{1000} = 0.2 \]\[ P(B) = \frac{700}{1000} = 0.7 \]Thus,\[ P(A|B) = \frac{0.2}{0.7} = \frac{2}{7} \]

Step 3: Calculate Probability for Part (b)

For part (b), we find the probability that a student owns a sound system given they own a car. This is given by \( P(B|A) \), where \( B \) is owning a sound system. The formula is:\[P(B|A) = \frac{P(A \cap B)}{P(A)}\]Similarly:\[ P(A \cap B) = \frac{200}{1000} = 0.2 \]\[ P(A) = \frac{450}{1000} = 0.45 \]So, the probability is:\[ P(B|A) = \frac{0.2}{0.45} = \frac{4}{9} \]

Key Concepts

Survey Data Analysis

Survey data analysis involves understanding and interpreting data collected from a group of respondents to draw meaningful insights. In this exercise, we examined a survey involving 1000 college students. The survey gathered data on students who own cars, sound systems, or both.

We are particularly interested in three numbers from this survey:

• 450 students own cars.
• 700 students own sound systems.
• 200 students own both a car and a sound system.

Analyzing survey data like this allows us to explore relationships between different sets, such as car and sound system ownership, and to calculate useful probabilities.

Probability Calculation

Probability calculation is a crucial aspect of understanding how likely an event is to occur. It provides a way to quantify uncertainty. In the context of our survey, we need to calculate conditional probabilities.

The concept of conditional probability helps determine how likely an event is to happen given that another event has already occurred. For example, finding the probability that a student owns a car given they own a sound system, which is represented by the notation \( P(A|B) \). The formula is:

• \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

In the solution, for Part (a), we calculated:

• \( P(A \cap B) = 0.2 \)
• \( P(B) = 0.7 \)
• This leads to \( P(A|B) = \frac{2}{7} \).

For Part (b), finding the probability a student owns a sound system given they own a car, uses:

• \( P(B|A) = \frac{P(A \cap B)}{P(A)} \)

Calculating:

• \( P(A) = 0.45 \)
• This leads to \( P(B|A) = \frac{4}{9} \).

Set Theory

Set theory is a fundamental concept in mathematics that deals with the collection of objects, known as sets. In probability, set theory is used to define events and their relationships.

In our problem, we deal with three important sets:

• Set \( A \) represents students owning cars.
• Set \( B \) represents students owning sound systems.
• The intersection of Set \( A \) and Set \( B \), denoted \( A \cap B \), includes students owning both a car and a sound system.

These sets are crucial to understand when calculating probabilities because they represent the different combinations of ownership among students.

Set operations like intersections (\( A \cap B \)) allow us to find probabilities of joint events. Understanding these relationships helps make informed decisions based on data, such as evaluating the impact of owning one item on the likelihood of owning another.