/*! 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 33 Jane Doe claims to possess extra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Jane Doe claims to possess extrasensory perception (ESP). She says she can guess more often than not the outcome of a flip of a balanced coin in another room. In an experiment, a coin is flipped three times. If she does not actually have ESP, find the probability distribution of the number of her correct guesses. a. Do this by constructing a sample space, finding the probability for each point, and using them to construct the probability distribution. b. Do this using the formula for the binomial distribution.

Short Answer

Expert verified
The probability distribution is: P(X=0)=1/8, P(X=1)=3/8, P(X=2)=3/8, P(X=3)=1/8.

Step by step solution

01

Define the Problem

Jane claims to have ESP and predict coin flips. We'll analyze this claim by evaluating the probability distribution of her correct guesses if she has no ESP. This means we will assume she guesses randomly.
02

Construct the Sample Space

For three independent coin flips, each with two possible outcomes (heads or tails), there are \[2^3 = 8\] total possible outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
03

Determine Probability of Each Outcome

For each coin flip, the probability of guessing correctly is \(\frac{1}{2}\). Therefore, the probability of each sequence of outcomes is \[\left(\frac{1}{2}\right)^3 = \frac{1}{8}\].
04

Count Winning Outcomes

For 0 correct guesses: none, for 1 correct guess: HTT, THT, TTH (3 outcomes), for 2 correct guesses: HHT, HTH, THH (3 outcomes), for 3 correct guesses: HHH (1 outcome).
05

Construct Probability Distribution

The probability distribution can be described as follows:- Probability of 0 correct guesses: \( P(X=0) = \frac{1}{8}\) (since only TTT).- Probability of 1 correct guess: \( P(X=1) = 3 \times \frac{1}{8} = \frac{3}{8}\).- Probability of 2 correct guesses: \( P(X=2) = 3 \times \frac{1}{8} = \frac{3}{8}\).- Probability of 3 correct guesses: \( P(X=3) = \frac{1}{8}\).
06

Use Binomial Distribution Formula

The binomial distribution formula is given by \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 3 \) (number of trials), \( p = \frac{1}{2} \) (success probability), and \( k \) is the number of correct guesses.- For \( k = 0 \), \( P(X=0) = \binom{3}{0} \left(\frac{1}{2}\right)^{0} \left(\frac{1}{2}\right)^{3} = \frac{1}{8} \).- For \( k = 1 \), \( P(X=1) = \binom{3}{1} \left(\frac{1}{2}\right)^{1} \left(\frac{1}{2}\right)^{2} = \frac{3}{8} \).- For \( k = 2 \), \( P(X=2) = \binom{3}{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{1} = \frac{3}{8} \).- For \( k = 3 \), \( P(X=3) = \binom{3}{3} \left(\frac{1}{2}\right)^{3} \left(\frac{1}{2}\right)^{0} = \frac{1}{8} \).

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.

Binomial Distribution
The binomial distribution is a key concept in probability theory that describes the number of successes in a fixed number of independent Bernoulli trials. Each trial has two possible outcomes: success or failure. In this context, guessing the correct side of a coin flip is considered a trial.

Here, Jane Doe's claim to guess the outcomes of coin flips can be modeled using a binomial distribution if we assume she doesn't possess extrasensory perception (ESP). The number of trials () is equal to the number of times the coin is flipped (3 flips in the exercise). Each trial is independent, and the probability of success (guessing correctly) is 0.5, since the coin is fair. The formula for the binomial distribution is given by:
  • \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]
This allows us to compute the probabilities for Jane guessing correctly 0, 1, 2, or 3 times.

Understanding the binomial distribution helps in analyzing the claims of others like Jane, as it provides a structured way to calculate the likelihood of different scenarios in experiments based on random processes.
Sample Space
Understanding the sample space is crucial when dealing with probability problems. The sample space of an experiment is the set of all possible outcomes. In Jane's coin-flipping exercise, each flip can result in either heads (H) or tails (T).

For three flips, the sample space consists of all combinations of these outcomes. By calculating \(2^3\), we see there are 8 possible outcomes.

Here's the list of possible outcomes:
  • HHH
  • HHT
  • HTH
  • THH
  • HTT
  • THT
  • TTH
  • TTT
Each outcome has an equal probability of occurring, which is key to determining individual probabilities. Recognizing and constructing the sample space lays the foundation for understanding any probability distribution associated with random trials.
Probability Theory
Probability theory forms the backbone of understanding how likely events are to occur. It is essential when analyzing claims like Jane's. Probability is quantified between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

In probability theory, calculating the probability of a specific event involves understanding the ratio of favorable outcomes to the total number of possible outcomes. Each possible outcome in the sample space is equally likely when dealing with fair coins, which means we can analyze Jane's ability to guess correctly without bias.

Let's breakdown this concept further:
  • The probability of no correct guesses (zero heads): When Jane guesses all three flips wrong, the only favorable outcome is TTT, leading to a probability of \( \frac{1}{8} \).
  • The probability of guessing all correctly (three heads): Only HHH qualifies, and it also has a probability of \( \frac{1}{8} \).
Understanding these probabilities is a fundamental part of probability theory, allowing us to construct a probability distribution and analyze what we might expect if Jane is indeed guessing randomly.

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

In \(2008,\) the murder rates (per 100,000 residents) for the 50 states and the District of Columbia (D.C.) had a mean of 5.39 and a standard deviation of 4.434 (Statistical Abstract of the United States). a. D.C. had a murder rate of 31.4 . Find its \(z\) -score. If the distribution were roughly normal, would this be unusually high? Explain. b. Based on the mean and standard deviation, do you think that the distribution of murder rates is approximately normal? Why or why not?

Binomial needs fixed \(n\) For the binomial distribution, the number of trials \(n\) is a fixed number. Let \(X\) denote the number of girls in a randomly selected family in Canada that has three children. Let \(Y\) denote the number of girls in a randomly selected family in Canada (that is, the number of children could be any number). A binomial distribution approximates well the probability distribution for one of \(X\) and \(Y,\) but not for the other. a. Explain why. b. Identify the case for which the binomial applies, and identify \(n\) and \(p\).

The state of Ohio has several statewide lottery options. One is the Pick 3 game in which you pick one of the 1000 three-digit numbers between 000 and 999\. The lottery selects a three-digit number at random. With a bet of \(\$ 1,\) you win \(\$ 500\) if your number is selected and nothing (\$0) otherwise. (Many states have a very similar type of lottery.) (Source: Background information from www.ohiolottery.com.) a. With a single \(\$ 1\) bet, what is the probability that you win \(\$ 500 ?\) b. Let \(X\) denote your winnings for a \(\$ 1\) bet, so \(x=\$ 0\) or \(x=\$ 500\). Construct the probability distribution for \(X\). c. Show that the mean of the distribution equals 0.50 , corresponding to an expected return of 50 cents for the dollar paid to play. Interpret the mean. d. In Ohio's Pick 4 lottery, you pick one of the 10,000 four-digit numbers between 0000 and 9999 and (with a \(\$ 1\) bet ) win \(\$ 5000\) if you get it correct. In terms of your expected winnings, with which game are you better off - playing Pick 4 , or playing Pick 3 in which you win \(\$ 500\) for a correct choice of a three-digit number? Justify your answer.

Some companies, such as DemandTec, have developed software to help retail chains set prices that optimize their profits. An Associated Press story (April 28,2007 ) about this software described a case in which a retail chain sold three similar power drills: one for \(\$ 90\), a better one for \(\$ 120\), and a top-tier one for \(\$ 130\). Software predicted that by selling the middle-priced drill for only \(\$ 110,\) the cheaper drill would seem less a bargain and more people would buy the middle-price drill. a. For the original pricing, suppose \(50 \%\) of sales were for the \(\$ 90\) drill, \(20 \%\) for the \(\$ 120\) drill, and \(30 \%\) for the \(\$ 130\) drill. Construct the probability distribution of \(X=\) selling price for the sale of a drill, and find its mean and interpret, b. For the new pricing, suppose \(30 \%\) of sales were for the \(\$ 90\) drill, \(40 \%\) for the \(\$ 110\) drill, and \(30 \%\) for the \(\$ 130\) drill. Is the mean of the probability distribution of selling price higher with this new pricing strategy? Explain.

SAT math scores follow a normal distribution with an approximate \(\mu=500\) and \(\sigma=100\). Also ACT math scores follow a normal distrubution with an approximate \(\mu=21\) and \(\sigma=4.7\). You are an admissions officer at a university and have room to admit one more student for the upcoming year. Joe scored 600 on the SAT math exam, and Kate scored 25 on the ACT math exam. If you were going to base your decision solely on their performances on the exams, which student should you admit? Explain.

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.