/*! 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 6 Fill in the blanks. If the occ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Fill in the blanks. If the occurrence of one event has no effect on the occurrence of a second event, then the events are ______.

Short Answer

Expert verified
Independent

Step by step solution

01

Understanding the Scenario

In a probability scenario, if the occurrence of one event does not influence the occurrence of a second event, those events are called 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 91Ó°ÊÓ!

Key Concepts

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

Understanding Probability
Probability is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, inclusive, where 0 indicates an impossibility of the event, and 1 indicates certainty. The higher the probability of an event, the more likely it is to occur.

For example, when flipping a fair coin, there are two possible outcomes - heads or tails. Each of these outcomes has an equal probability of occurring, which is expressed numerically as 0.5 or a 50% chance. In mathematical terms, this is represented as:
\[ P(\text{Heads}) = P(\text{Tails}) = 0.5 \]
where P(event) denotes the probability of an event. Understanding probability is crucial as it provides a way to predict the chances of various outcomes in different scenarios, from simple games of chance to complex real-world situations.
Navigating Probability Scenarios
Probability scenarios involve situations in which multiple events can occur, with each event having its own set of probabilities. These scenarios can be simple, such as rolling a single die, or complex, involving multiple stages and choices. To navigate these, it's important to consider the total number of possible outcomes and the number of outcomes that correspond to the event of interest.

For instance, when rolling a six-sided die, you have six possible outcomes (1 through 6). The probability of rolling a three is:
\[ P(\text{Rolling a Three}) = \frac{1}{6} \]
because only one outcome (rolling a three) matches the event of interest out of six possible outcomes. In more complex scenarios, one might have to use combinations or permutations to determine the number of favorable outcomes.
The Concept of Independent Probability
Independent probability refers to the situation where the occurrence of one event does not influence the occurrence of another event. In other words, the events have no impact on each other's outcomes. This concept is foundational in probability theory as it dictates how probabilities of multiple events can be calculated together.

When dealing with independent events, the probability of both events occurring is calculated by multiplying the probabilities of each event. For example, consider rolling a die and flipping a coin simultaneously. Rolling a five (event A) and flipping heads (event B) are independent events, so their combined probability is:
\[ P(A \text{ and } B) = P(A) \times P(B) = \frac{1}{6} \times 0.5 = \frac{1}{12} \]
This formula allows us to handle complex scenes where multiple independent events must be considered.

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

Simplifying a Difference Quotient In Exercises \(67-72\) , simplify the difference quotient, using the Binomial Theorem if necessary. $$\frac{f(x+h)-f(x)}{b} \quad$$ Difference quotient $$f(x)=x^{4}$$

Probability In Exercises \(85-88,\) consider \(n\) independent trials of an experiment in which each trial has two possible outcomes: "success" or "failure.". The probability of a success on each trial is \(p,\) and the probability of a failure is \(q=1-p .\) In this context, the term \(_{n} C_{k} p^{k} q^{n-k}\) in the expansion of \((p+q)^{n}\) gives the probability of \(k\) successes in the \(n\) trials of the experiment. To find the probability that the sales representative in Exercise 87 makes four sales when the probability of a sale with any one customer is \(\frac{1}{2},\) evaluate the term 8 $$_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{4}$$ in the expansion of \(\left(\frac{1}{2}+\frac{1}{2}\right)^{8}\)

Assume that the probability of the birth of a child of a particular sex is 50\(\% .\) In a family with four children, what is the probability that (a) all the children are boys, (b) all the children are the same sex, and (c) there is at least one boy?

Approximation In Exercises \(79-82,\) use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \((1.02)^{8}=(1+0.02)^{8}\) $$=1+8(0.02)+28(0.02)^{2}+\cdots+(0.02)^{8}$$ $$(2.99)^{12}$$

Simplifying a Difference Quotient In Exercises \(67-72\) , simplify the difference quotient, using the Binomial Theorem if necessary. $$\frac{f(x+h)-f(x)}{b} \quad$$ Difference quotient $$f(x)=\sqrt{x}$$

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.