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Chapter 4: Problem 36

In a large city, 15,000 workers lost their jobs last year. Of them, 7400 lost their jobs because their companies closed down or moved, 4600 lost their jobs due to insufficient work, and the remainder lost their jobs because their positions were abolished. If one of these 15,000 workers is selected at random, find the probability that this worker lost his or her job a. because the company closed down or moved b. due to insufficient work c. because the position was abolished Do these probabilities add up to \(1.0 ?\) If so, why?

Short Answer

Expert verified
The probability a worker lost their job because the company closed down or moved is 0.4933, the probability a worker lost his job due to insufficient work is 0.3067, and the probability a worker lost their job because their position was abolished is 0.2. The sum of these probabilities is 1.0, which is the expected total since they represent all possible outcomes.

Step by step solution

01

Calculate Probability for Each Reason

To calculate the probability we take the number of individuals affected by a specific event and divide it by the total number of individuals. \n\nThe probability a worker lost his job because the company closed down or moved \(P(A) = \frac{7400}{15000} = 0.4933\). \n\nThe probability a worker lost his job due to insufficient work \(P(B) = \frac{4600}{15000} = 0.3067\). \n\nIn order to find out how many workers lost their jobs because their positions were abolished, subtract the number of workers who lost their jobs due to the other two reasons from the total number of workers who lost their jobs. Therefore, \(15000 鈥� 7400 - 4600 = 3000\). Hence, the probability a worker lost his job because their position was abolished \(P(C) = \frac{3000}{15000} = 0.2\).
02

Check If Probabilities Add Up To 1.0

Now add all individual probabilities \(P(A) + P(B) + P(C) = 0.4933 + 0.3067 + 0.2 = 1\).\nHence, these probabilities do add up to 1 because these are all possible outcomes in this scenario. As per the rules of probability, the sum of probabilities for all possible outcomes should always equal 1

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Key Concepts

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

Random Selection
Random selection is a process used to ensure fairness in choosing one individual from a group. In the context of probability, it means picking one item randomly from a set without bias or rules dictating the choice. Imagine 15,000 workers who lost their jobs for various reasons. If a worker is chosen at random, every worker should have an equal chance of being selected.

Random selection is like drawing names from a hat. Each name inside has the same chance of being pulled out. This randomness ensures every possible outcome is equally likely initially. In situations involving probabilities, like our exercise about job loss reasons, random selection is key to fairly assess the situation without favoring any particular outcome.
Probability Calculation
Calculating probability is all about understanding how likely an event is to happen. For any given event, its probability is calculated using the formula:
  • Probability (P) = \( \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
In the job loss scenario, there are several specific reasons why workers lost their jobs. To find the probability for each reason, count how many workers lost their jobs for that reason and divide by the total number of workers affected.
  • For instance, for a company closing or moving, with 7,400 workers: \[ P(\text{closed/moved}) = \frac{7400}{15000} = 0.4933 \]
This calculation gives a clear picture of how probable each reason was for causing job loss. Understanding this helps in analyzing real-world situations based on quantitative data.
Sum of Probabilities
In probability theory, the sum of probabilities for all possible outcomes in an experiment should equal to 1. This concept is known as the probability axiom. It makes sure all potential scenarios are accounted for and no probabilities are missed.

In the given exercise, we calculate probabilities for three mutually exclusive reasons of job loss and add them up:
  • Closed/Move: \( 0.4933 \)
  • Insufficient Work: \( 0.3067 \)
  • Position Abolished: \( 0.2 \)
Adding these probabilities yields:
  • \( 0.4933 + 0.3067 + 0.2 = 1 \)
This sum equaling 1 confirms that the probabilities are accurately calculated and represent all possible outcomes in this scenario.
Job Loss Reasons
Understanding why something happens can be essential for predicting future occurrences or taking preventative actions. In the context of the exercise, job loss reasons are classified into three main categories:
  • Company Closed or Moved: This indicates a significant structural change in the business environment, where some companies either close their operations or move to a different location.
  • Insufficient Work: A typical scenario where there are not enough jobs or assignments available to keep everyone employed. This can happen due to economic downturns reducing demand.
  • Position Abolished: Sometimes, certain job roles become redundant due to technological advancements or restructuring in the company, leading to their elimination.
By analyzing the reason for each job loss, we can gain insights into the economic and operational health of the job market. This understanding is crucial for both employees seeking job security and employers planning strategic changes in hiring and operations.

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Most popular questions from this chapter

What is meant by the joint probability of two or more events? Give one example.

Twenty percent of a town's voters favor letting a major discount store move into their neighborhood, \(63 \%\) are against it, and \(17 \%\) are indifferent. What is the probability that a randomly selected voter from this town will either be against it or be indifferent? Explain why this probability is not equal to \(1.0\).

There is an area of free (but illegal) parking near an inner-city sports arena. The probability that a car parked in this area will be ticketed by police is \(.35\), that the car will be vandalized is \(.15\), and that it will be ticketed and vandalized is \(.10 .\) Find the probability that a car parked in this area will be ticketed or vandalized.

Suppose that \(20 \%\) of all adults in a small town live alone, and \(8 \%\) of the adults live alone and have at least one pet. What is the probability that a randomly selected adult from this town has at least one pet given that this adult lives alone?

Refer to Exercise 4.52, which contains information on a July 21, 2009 www.HuffingtonPost.com survey that asked people to choose their favorite junk food from a list of choices. The following table contains results classified by gender. (Note: There are 4801 females and 3201 males.) $$ \begin{array}{lcc} \hline \text { Favorite Junk Food } & \text { Female } & \text { Male } \\ \hline \text { Chocolate } & 1518 & 531 \\ \text { Sugary candy } & 218 & 127 \\ \text { Ice cream } & 685 & 586 \\ \text { Fast food } & 312 & 463 \\ \text { Cookies } & 431 & 219 \\ \text { Chips } & 458 & 649 \\ \text { Cake } & 387 & 103 \\ \text { Pizza } & 792 & 523 \\ \hline \end{array} $$ a. Suppose that one person is selected at random from this sample of 8002 respondents. Find the following probabilities. i. Probability of the intersection of events female and ice cream. ii. Probability of the intersection of events male and pizza. b. Mention at least four other joint probabilities you can calculate for this table and then find their probabilities. You may draw a tree diagram to find these probabilities.
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