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Chapter 4: Problem 135

In a class of 35 students, 13 are seniors, 9 are juniors, 8 are sophomores, and 5 are freshmen. If one student is selected at random from this class, what is the probability that this student is a. a junior? b. a freshman?

Short Answer

Expert verified
a. The probability of a randomly selected student being a junior is \( \frac{9}{35} \); b. The probability of a randomly selected student being a freshman is \( \frac{5}{35} \).

Step by step solution

01

Understand the problem

There are 35 students in total. Out of these, there are 9 juniors and 5 freshmen. The task is to find the probability of selecting a junior or a freshman out of these 35 students.
02

Calculate the probability of a junior

Probability is calculated by dividing the number of successful outcomes by the total number of outcomes. Therefore, the probability \( P \) of selecting a junior is equal to the number of juniors divided by the total number of students. \( P(\mathrm{Junior}) = \frac{9}{35} \).
03

Calculate the probability of a freshman

Similarly, the probability of selecting a freshman is equal to the number of freshmen divided by the total number of students. \( P(\mathrm{Freshman}) = \frac{5}{35} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Selection in Probability
Random selection is a fundamental concept in probability. It means choosing an item, person, or outcome without any preference or bias. This ensures that every option has an equal chance of being picked. In our classroom scenario, each student has an equal chance of being selected when we choose randomly.
  • Every one of the 35 students in the class has an equal opportunity to be selected.
  • No favoritism or external criteria influence the selection process.
This equal chance is crucial in calculating probabilities because it sets the stage for fair decision-making. If selection isn鈥檛 random, our probability calculations would not be valid since some outcomes would inherently be more likely than others.
Probability Calculation
The core idea of probability calculation is determining the likelihood of a specific outcome. It's represented as a fraction where the numerator is the number of favorable outcomes and the denominator is the total number of possible outcomes.
For example, in our task, we have two separate queries:
  • The probability of selecting a junior student is determined by dividing the number of juniors (9) by the total number of students (35): \( P(\text{Junior}) = \frac{9}{35} \).
  • The probability of selecting a freshman is calculated the same way, using the number of freshmen (5): \( P(\text{Freshman}) = \frac{5}{35} \).
Understanding this formula means you've grasped the essence of calculating probabilities - simply put, favorable outcomes over total outcomes.
Student Classification
In probability and data analysis, classification is an effective way to organize information. Here, students are classified based on their academic year: seniors, juniors, sophomores, and freshmen.
This classification helps in:
  • Identifying the count of each group, such as recognizing there are 13 seniors, 9 juniors, 8 sophomores, and 5 freshmen in the class.
  • Simplifying the process of calculating probabilities by focusing on each specific category.
Sound classification ensures efficient data handling and is critical for accurate probability assessments. It allows us to focus on precise subsets of data, making it easier to solve problems by breaking them down into smaller, more manageable parts.

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Most popular questions from this chapter

A production system has two production lines; each production line performs a two-part process, and each process is completed by a different machine. Thus, there are four machines, which we can identify as two first-level machines and two second-level machines. Each of the first-level machines works properly \(98 \%\) of the time, and each of the second-level machines works properly \(96 \%\) of the time. All four machines are independent in regard to working properly or breaking down. Two products enter this production system, one in each production line. a. Find the probability that both products successfully complete the two-part process (i.e., all four machines are working properly). b. Find the probability that neither product successfully completes the two- part process (i.e., at least one of the machines in each production line is not working properly).

A consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} \text { 30 Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \\ \hline \end{array} $$ If one flight is selected at random from these 1700 flights, find the following probabilities. a. \(P\) (more than 1 hour late or airline \(\mathrm{A}\) ) b. \(P(\) airline \(B\) or less than 30 minutes late) c. \(P(\) airline A or airline \(\mathrm{B}\) )

A company employs a total of 16 workers. The management has asked these employees to select 2 workers who will negotiate a new contract with management. The employees have decided to select the 2 workers randomly. How many total selections are possible? Considering that the order of selection is important, find the number of permutations.

In a political science class of 35 students, 21 favor abolishing the electoral college and thus electing the President of the United States by popular vote. If two students are selected at random from this class, what is the probability that both of them favor abolition of the electoral college? Draw a tree diagram for this problem.

A gambler has four cards - two diamonds and two clubs. The gambler proposes the following game to you: You will leave the room and the gambler will put the cards face down on a table. When you return to the room, you will pick two cards at random. You will win $$\$ 10$$ if both cards are diamonds, you will win $$\$ 10$$ if both are clubs, and for any other outcome you will lose $$\$ 10$$. Assuming that there is no cheating, should you accept this proposition? Support your answer by calculating your probability of winning $$\$ 10$$.
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