Question

In a certain Algebra and Trigonometry class, there are 18 freshmen and 15 sophomores. Of the 18 freshmen, 10 are male, and of the 15 sophomores, 8 are male. Find the probability that a randomly selected student is: (a) A freshman or female (b) A sophomore or male

Short answer

Probability for part (a) is \(\frac{25}{33}\). Probability for part (b) is \(\frac{25}{33}\).

Step-by-step solution

- Determine Total Number of Students

Add the number of freshmen and sophomores together: \[ 18 \text{ (freshmen)} + 15 \text{ (sophomores)} = 33 \text{ (total students)} \]

- Find the Probability of a Freshman

Calculate the probability of selecting a freshman: \[ P(\text{Freshman}) = \frac{18}{33} \]

- Find the Number of Female Students

Calculate the number of female students: \[ 18 \text{ (freshmen)} - 10 \text{ (male freshmen)} = 8 \text{ (female freshmen)} \] \[ 15 \text{ (sophomores)} - 8 \text{ (male sophomores)} = 7 \text{ (female sophomores)} \] Total number of female students = \[ 8 + 7 = 15 \]

- Find the Probability of a Female

Calculate the probability of selecting a female student: \[ P(\text{Female}) = \frac{15}{33} \]

- Calculate the Probability of Either Freshman or Female

Add the probabilities of selecting a freshman or a female student, but subtract the probability of being both to avoid double counting: \[ P(\text{Freshman or Female}) = P(\text{Freshman}) + P(\text{Female}) - P(\text{Freshman and Female}) \] \[ P(\text{Freshman}) = \frac{18}{33} \] \[ P(\text{Female}) = \frac{15}{33} \] Number of female freshmen = 8, hence: \[ P(\text{Freshman and Female}) = \frac{8}{33} \] Therefore: \[ P(\text{Freshman or Female}) = \frac{18}{33} + \frac{15}{33} - \frac{8}{33} = \frac{25}{33} \]

- Find the Probability of a Sophomore

Calculate the probability of selecting a sophomore: \[ P(\text{Sophomore}) = \frac{15}{33} \]

- Find the Probability of a Male

Calculate the probability of selecting a male student: \[ 10 \text{ (male freshmen)} + 8 \text{ (male sophomores)} = 18 \text{ (total male students)} \] \[ P(\text{Male}) = \frac{18}{33} \]

- Calculate the Probability of Either Sophomore or Male

Add the probabilities of selecting a sophomore or a male student, but subtract the probability of being both to avoid double counting: \[ P(\text{Sophomore or Male}) = P(\text{Sophomore}) + P(\text{Male}) - P(\text{Sophomore and Male}) \] \[ P(\text{Sophomore}) = \frac{15}{33} \] \[ P(\text{Male}) = \frac{18}{33} \] Number of male sophomores = 8, hence: \[ P(\text{Sophomore and Male}) = \frac{8}{33} \] Therefore: \[ P(\text{Sophomore or Male}) = \frac{15}{33} + \frac{18}{33} - \frac{8}{33} = \frac{25}{33} \]

Key concepts

Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has already occurred. In our problem, calculating the probability of a student being a freshman or female, or a sophomore or male, involves this concept substantially.
To find the probability of either one of two events happening, we need to add the probabilities of the individual events and then subtract the probability of both occurring together.
For instance, to calculate the probability of a student being a freshman or a female, we first find the probability of selecting a freshman and a female, then use the formula:
\[P(\text{Freshman or Female}) = P(\text{Freshman}) + P(\text{Female}) - P(\text{Freshman and Female})\]
Simultaneously, to make accurate predictions about the entire sample space, you need to consider the overlapping probabilities carefully. This avoids counting the same event multiple times, hence adjusting probabilities accurately.

Combinations

Combinations help us determine how many ways we can choose a group of items from a larger set, without considering the order. In probability problems like the one given, combinations are implicit when we consider groups of students.
For example, determining the number of male or female students from the freshman and sophomore groups involves counting combinations.
The total number of students provides the framework, and the numbers of males and females are subsets of this larger set. Although combinations are more explicit in more complex scenarios where order doesn’t matter, the fundamental concept aids in formulating and understanding probability calculations effectively.
Combinations are fundamental in probability since they help to build the sample space—the set of all possible outcomes.

Sample Space

The sample space in probability is the set of all possible outcomes of a random experiment. For our problem, the sample space is the total number of students.
To determine probabilities accurately, it's essential to identify and understand the sample space first.
In the current problem, we have:

• Total students: 33
• Freshmen: 18
• Sophomores: 15
• Males: 18
• Females: 15

All calculations like finding the probability of a freshman or female, or the probability of a sophomore or male, are based on this sample space.
By thoroughly understanding the sample space, we can ensure that all potential outcomes are considered, leading to accurate probability calculations.