/*! 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 18 Suppose two balanced coins are t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose two balanced coins are tossed and the upper faces are observed. a. List the sample points for this experiment. b. Assign a reasonable probability to each sample point. (Are the sample points equally likely?) c. Let \(A\) denote the event that exactly one head is observed and \(B\) the event that at least one head is observed. List the sample points in both \(A\) and \(B\). d. From your answer to part \((c),\) find \(P(A), P(B), P(A \cap B), P(A \cup B),\) and \(P(\bar{A} \cup B)\)

Short Answer

Expert verified
P(A) = 1/2, P(B) = 3/4, P(A ∩ B) = 1/2, P(A ∪ B) = 3/4, P(Ā ∪ B) = 1.

Step by step solution

01

List Sample Points

When two coins are tossed, each coin can be either Heads (H) or Tails (T). So, the combinations of the results (sample points) for two coins are: HH, HT, TH, and TT.
02

Assign Probabilities

Each sample point (HH, HT, TH, TT) has an equal probability of occurring because the coins are balanced. There are a total of 4 sample points, so each has a probability of \( \frac{1}{4} \).
03

Identify Event A (Exactly One Head)

For event A to occur, we need exactly one head, which can be either HT or TH. Therefore, the sample points in event A are HT and TH.
04

Identify Event B (At Least One Head)

For event B to occur, we need at least one head, which includes the sample points HH, HT, and TH.
05

Calculate Probabilities

- \( P(A) \): Event A contains HT and TH, so \( P(A) = \frac{2}{4} = \frac{1}{2} \).- \( P(B) \): Event B contains HH, HT, and TH, so \( P(B) = \frac{3}{4} \).- \( P(A \cap B) \): Both A and B include sample points HT and TH, so \( P(A \cap B) = \frac{2}{4} = \frac{1}{2} \).- \( P(A \cup B) \): Union of A and B (all sample points with at least one head) gives HH, HT, TH, so \( P(A \cup B) = \frac{3}{4} \).- \( P(\bar{A} \cup B) \): \( \bar{A} \) is event not A, which includes HH and TT. Union with B results in HH, HT, TH, TT, so \( P(\bar{A} \cup B) = 1 \).

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 Sample Space
In probability theory, the sample space refers to the set of all possible outcomes of a particular experiment. When we're considering an experiment like tossing two balanced coins, we're looking at all the different ways the coins can land. Each possible combination of outcomes is called a sample point.
For our coin-tossing experiment, the sample space includes these possibilities:
  • Both coins show heads (HH)
  • The first coin shows heads while the second shows tails (HT)
  • The first coin shows tails while the second shows heads (TH)
  • Both coins show tails (TT)

Understanding the sample space is a crucial first step in calculating probabilities, as it represents the full set of results that we will evaluate.
Event Probability Explained
Probability helps us find the likelihood of specific events occurring. In our coin-tossing scenario, when the coins are balanced, each outcome is equally likely.
With four possible outcomes (HH, HT, TH, TT) and equal probability distribution, each sample point has a probability of occurring of \( \frac{1}{4} \). This is because there are four total sample points, and the probability of each event happening (assuming all are equally likely) is the inverse of the total number of outcomes.
In terms of events such as "exactly one head," or "at least one head," we express the event probability as a fraction of the sample points. The probability of event \(A\) (exactly one head) is computed by considering the sample points that meet this condition: HT and TH. Hence, \(P(A) = \frac{2}{4} = \frac{1}{2}\). Similarly, the probability of event \(B\) (at least one head) is derived from HT, TH, and HH, resulting in \(P(B) = \frac{3}{4}\). This segmentation helps in analyzing events based on their specific characteristics.
Independent Events in Probability
Events are considered independent when the occurrence of one event does not affect the probability of the other. In this coin-tossing scenario, the toss of one coin does not impact the toss of the other coin. This means each coin toss is an independent event.
The probability of flipping one head or tail remains \( \frac{1}{2} \), regardless of previous or concurrent outcomes. Recognizing events as independent simplifies the calculation of joint probabilities, as individual probabilities can be multiplied. Even when calculating \(P(A \cap B)\) or the probability of intersections in independent events, the independence allows treating each event distinctly while preserving their unique likelihoods.
A Glance at Combinatorics
Combinatorics is a branch of mathematics that studies the combination of objects. It helps us count different possible outcomes and solve problems related to probability. In our coin-tossing example, we use combinatorics to determine all potential outcomes.
The basic combinatorial method is listing possibilities or arranging, depending on the number of objects. Here, the objects are two coins, and each coin has two positions, resulting in © combinations. This results in the four outcomes we have analyzed as our sample space: HH, HT, TH, and TT.
Combinatorics also lends itself to formulating probabilities efficiently when more complex situations arise, assisting in calculating and analyzing interesting events and outcomes systematically.

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

A business office orders paper supplies from one of three vendors, \(V_{1}, V_{2},\) or \(V_{3} .\) Orders are to be placed on two successive days, one order per day. Thus, \(\left(V_{2}, V_{3}\right)\) might denote that vendor \(V_{2}\) gets the order on the first day and vendor \(V_{3}\) gets the order on the second day. a. List the sample points in this experiment of ordering paper on two successive days. b. Assume the vendors are selected at random each day and assign a probability to each sample point. c. Let \(A\) denote the event that the same vendor gets both orders and \(B\) the event that \(V_{2}\) gets at least one order. Find \(P(A), P(B), P(A \cup B),\) and \(P(A \cap B)\) by summing the probabilities of the sample points in these events.

Five identical bowls are labeled \(1,2,3,4,\) and \(5 .\) Bowl \(i\) contains \(i\) white and \(5-i\) black balls, with \(i=1,2 \ldots, 5 .\) A bowl is randomly selected and two balls are randomly selected (without replacement) from the contents of the bowl. a. What is the probability that both balls selected are white? b. Given that both balls selected are white, what is the probability that bowl 3 was selected?

According to Webster's New Collegiate Dictionary, a divining rod is "a forked rod believed to indicate [divine] the presence of water or minerals by dipping downward when held over a vein." To test the claims of a divining rod expert, skeptics bury four cans in the ground, two empty and two filled with water. The expert is led to the four cans and told that two contain water. He uses the divining rod to test each of the four cans and decide which two contain water. a. List the sample space for this experiment. b. If the divining rod is completely useless for locating water, what is the probability that the expert will correctly identify (by guessing) both of the cans containing water?

An assembly operation in a manufacturing plant requires three steps that can be performed in any sequence. How many different ways can the assembly be performed?

The symmetric difference between two events \(A\) and \(B\) is the set of all sample points that are in exactly one of the sets and is often denoted \(A \Delta B\). Note that \(A \Delta B=(A \cap B) \cup(\bar{A} \cap B)\). Prove that \(P(A \Delta B)=P(A)+P(B)-2 P(A \cap B)\)

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.