Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Suppose \(P(A)=.1\) and \(P(B)=.5 .\) $$\text { If } P(A \cap B)=0, \text { are } A \text { and } B \text { independent? }$$

Short Answer

Expert verified
Answer: No, events A and B are not independent.

Step by step solution

01

Understand the given probabilities

We are given the probabilities of events A and B as: - \(P(A) = 0.1\) - \(P(B) = 0.5\) - \(P(A \cap B) = 0\)
02

Calculate the product of individual probabilities

Compute the product of the probabilities of events A and B: \(P(A) * P(B) = 0.1 * 0.5 = 0.05\)
03

Compare the intersection probability and product to determine independence

To verify if events A and B are independent, compare the probability of their intersection, \(P(A \cap B)\), to the product of their individual probabilities, \(P(A) * P(B)\): - \(P(A \cap B) = 0\) - \(P(A) * P(B) = 0.05\) Since \(P(A \cap B) \neq P(A) * P(B)\), events A and B are not independent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Event Independence
In probability theory, independence is a key concept used to understand the relationship between two events. Two events, say A and B, are independent if the occurrence of one event does not affect the probability of the other occurring.
This implies that learning whether A occurred gives no information about whether B occurred and vice versa. This is a fundamental concept when dealing with probabilities as it simplifies calculating probabilities when events are independent.
  • If two events are independent, the probability of both occurring, or their intersection, can be found simply by multiplying the probabilities of each event occurring on its own. This is expressed as: \[ P(A \cap B) = P(A) \times P(B) \]
  • It is crucial to differentiate between independence and events being mutually exclusive, which means they cannot occur simultaneously. When events are mutually exclusive, the occurrence of one rules out the occurrence of the other, typically leading to a zero intersection probability. Independence, on the other hand, has nothing to do with mutual exclusivity.
Intersection Probability
Intersection probability helps determine how likely it is for two events to happen at the same time. For instance, if we want to know the probability of both A and B occurring, we refer to the intersection probability, \(P(A \cap B)\). This probability is crucial in determining whether two events are independent.
If the intersection probability aligns with the product of their individual probabilities, then the events are independent. Consequently, any deviation from this equivalence suggests that the events have some form of dependency.
However, if \(P(A \cap B) = 0\), it primarily indicates that the two events cannot occur together – they have no overlap in occurrence. This does not automatically signify independence. Instead, one should check the relationship between \(P(A \cap B)\) and \(P(A) \times P(B)\) to conclude on independence.
Probability Calculation
Probability calculation is a foundational aspect of understanding and predicting event outcomes. It's an essential tool in event analysis whether events are independent or not.
  • The first step is specifying the probability of each individual event. In our problem, we have \(P(A) = 0.1\) and \(P(B) = 0.5\).
  • When events are evaluated for independence, calculating their joint probability, \(P(A \cap B)\), is vital. In the given problem, this value is 0.
  • The comparison between the joint probability and the product of individual probabilities is crucial. For our example, \(P(A \cap B)\) is compared with \(P(A) \times P(B) = 0.05\).
The concept of comparing these values allows us to determine that if \(P(A \cap B)\) differs from the product \(P(A) \times P(B)\), the events are not independent.
This calculation not only helps in verifying independence but is also applicable across various probability scenarios, making it an invaluable skill.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Define the simple events for the experiments in Exercises \(16-20 .\) Three children are randomly selected and their gender is recorded.

When an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2}\) and \(S_{3}\), can occur, with \(P\left(S_{1}\right)=.2, P\left(S_{2}\right)=.5,\) and \(P\left(S_{3}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}, S_{2}\), or \(S_{3}\) has occurred are $$ P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3 $$ If event A is observed, use this information to find the probabilities in Exercises 4 -6. \(P\left(S_{2} \mid A\right)\)

The failure rate for a guided missile control system is 1 in \(1000 .\) Suppose that a duplicate, but completely independent, control system is installed in each missile so that, if the first fails, the second can take over. The reliability of a missile is the probability that it does not fail. What is the reliability of the modified missile?

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

A football team is known to run \(30 \%\) of its plays to the left and \(70 \%\) to the right. A linebacker on an opposing team notices that, when plays go to the right, the right guard shifts his stance most of the time \((80 \%)\) and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance \(90 \%\) of the time and the shift stance the remaining \(10 \%\). On a particular play, the linebacker notes that the guard takes a balanced stance. a. What is the probability that the play will go to the left? b. What is the probability that the play will go to the right? c. If you were the linebacker, which direction would you prepare to defend if you saw the balanced stance?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free