/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 When is the following addition r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When is the following addition rule used to find the probability of the union of two events \(A\) and \(B\) ? $$ P(A \text { or } B)=P(A)+P(B) $$ Give one example where you might use this formula.

Short Answer

Expert verified
The rule \(P(A \text { or } B) = P(A) + P(B)\) is used when events A and B are mutually exclusive. An example could be the probability of drawing a heart or diamond card from a deck, as these two events can't both happen simultaneously.

Step by step solution

01

Understand mutual exclusivity

Firstly, identify the concept of mutually exclusive events. These are events that cannot happen simultaneously. In other words, if one event happens, the other cannot.
02

Recognize the applicability of the rule

The next step is to understand that the addition rule \(P(A \text { or } B) = P(A) + P(B)\) is applied when the events, A and B, are mutually exclusive. This means, the rule is used when the intersection of A and B equals zero, or in probability notation, when \(P(A \cap B) = 0\).
03

Provide an example

Now, let's consider an example: Suppose we want to find the probability of drawing a heart or a diamond from a deck of playing cards. Since these two events are mutually exclusive (a card cannot be both a heart and a diamond at the same time), the probability is simply the sum of the separate probabilities. This is a situation where the formula \(P(A \text { or } B) = P(A) + P(B)\) could be used.

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 91Ó°ÊÓ!

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

According to a survey of 2000 home owners, 800 of them own homes with three bedrooms, and 600 of them own homes with four bedrooms. If one home owner is selected at random from these 2000 home owners, find the probability that this home owner owns a home that has three or four bedrooms. Explain why this probability is not equal to \(1.0 .\)

Two students are randomly selected from a statistics class, and it is observed whether or not they suffer from math anxiety. How many total outcomes are possible? Draw a tree diagram for this experiment.

Given that \(P(B \mid A)=.70\) and \(P(A\) and \(B)=.35\), find \(P(A)\).

A restaurant menu has four kinds of soups, eight kinds of main courses, five kinds of desserts, and six kinds of drinks. If a customer randomly selects one item from each of these four categories, how many different outcomes are possible?

A player plays a roulette game in a casino by betting on a single number each time. Because the wheel has 38 numbers, the probability that the player will win in a single play is \(1 / 38\). Note that each play of the game is independent of all previous plays. a. Find the probability that the player will win for the first time on the 10 th play. b. Find the probability that it takes the player more than 50 plays to win for the first time. c. A gambler claims that because he has 1 chance in 38 of winning each time he plays, he is certain to win at least once if he plays 38 times. Does this sound reasonable to you? Find the probability that he will win at least once in 38 plays.

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.