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Chapter 8: Problem 41

There are 12 signs of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagittarius, Capricorn, Aquarius, and Pisces. Each sign corresponds to a different calendar period of approximately 1 month. Assuming that a person is just as likely to be born under one sign as another, what is the probability that in a group of five people at least two of them a. Have the same sign? b. Were born under the sign of Aries?

Short Answer

Expert verified
a. The probability that at least 2 people have the same sign is: \(P\,(same)= 1-\frac{12*11*10*9*8}{12^5}\) b. The probability that at least two people were born under the sign of Aries is: \(P\,(2 \, Aries)=1-\left[\left(\frac{11}{12}\right)^5 + 5 \times \frac{1}{12} \times \left(\frac{11}{12}\right)^4\right]\)

Step by step solution

01

a. Probability that at least 2 people have the same sign

First, we find the complementary probability: the probability that all 5 people have different signs. There are 12 options for the first person, 11 options for the second (to not be the same sign as the first), 10 options for the third (to not be one of the first two), 9 for the fourth, and 8 for the last person. So, there are \(12*11*10*9*8\) different ways that all 5 people have unique signs. Since there are 12 signs and 5 people, each person could be born under any of the 12 signs, so there are a total of \(12^5\) outcomes. The probability that all of them have different signs is: \[ P\,(different)\,=\frac{12*11*10*9*8}{12^5} \] Now, to find the probability that at least 2 people have the same sign, we will subtract the above probability from 1: \[ P\,(same)\,=1-P\,(different)\,= 1-\frac{12*11*10*9*8}{12^5} \]
02

b. Probability that at least two people were born under the sign of Aries

Similar to (a), we will find the complementary probability: the probability that at most 1 person was born under the sign of Aries. There are 4 cases to consider here: 1. No one was born under Aries: probability = \(\left(\frac{11}{12}\right)^5\) 2. Exactly 1 person was born under Aries: probability = \(5 \times \left(\frac{1}{12}\right) \times \left(\frac{11}{12}\right)^4\) Then we will add those probabilities to find the complementary probability and subtract it from 1: \[ P\,(2 \, Aries)=1-\left[\left(\frac{11}{12}\right)^5 + 5 \times \frac{1}{12} \times \left(\frac{11}{12}\right)^4\right] \]

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Key Concepts

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

Zodiac Sign Probabilities
When we talk about zodiac sign probabilities, we're dealing with the chances of certain zodiac signs occurring within a set of individuals. In this exercise, we have a group of five people, and we want to calculate the probability that at least two of them have the same zodiac sign.

To make this concept more relatable, imagine you're at a party, and you began chatting with four other people about zodiac signs. The odds that at least two of you share the same sign can be surprising.

Understanding the Basics

In this scenario, we assume that a person is equally likely to be born under any of the 12 zodiac signs. This assumption gives each sign an equal probability of 1/12. When we have five people, each person has a 1/12 chance of being born under, for example, Aries or any given sign.

Application in the Exercise

In our exercise, finding this probability involves a two-part process. First, we calculate the probability of the opposite event 鈥� that all five people have different zodiac signs, which leads us directly to the concept of complementary probability that we'll discuss in the next section.
Complementary Probability
The complementary probability is essential when calculating the likelihood of an event by considering its opposite. In the zodiac sign example, instead of directly calculating the probability that at least two people share the same sign, we first determine the probability that everyone has a unique sign, and then take the complement of this probability.

How to Calculate the Complement

Let's break it down using our example. We find the number of ways each person can have a unique sign, and then place that over the total number of possible zodiac distributions for the five people. The complementary probability is then one minus this value, representing the primary event we're interested in 鈥� namely, that at least two people out of five share the same zodiac sign.

Importance in Problem-Solving

Using complementary probability can simplify calculations, especially when the event we're interested in has many possible outcomes, as it is the case with finding at least two people having the same sign among five. It helps avoid complex and lengthy calculations, making probability problems more manageable.
Probabilistic Outcomes
Talking about probabilistic outcomes refers to all the possible results that can emerge from a random event. It鈥檚 like imagining every possible way a situation could play out. In probability theory, we often use it to gauge the likelihood of different scenarios.

Analysis of the Scenario

For our zodiac example under part b, we analyzed two specific outcomes: no one being born under Aries and exactly one person being born under Aries. By calculating the probabilities for these specific outcomes, we got closer to understanding the overall probability space for this part of the problem.

Significance for the Student

A clear grip on probabilistic outcomes allows students to approach problems systematically by considering all different scenarios that can occur, and then adding or subtracting probabilities as needed. This is particularly helpful for complex problems, where direct computation isn鈥檛 feasible, and a more strategic approach is necessary.

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Most popular questions from this chapter

During the first year at a university that uses a 4 -point grading system, a freshman took ten 3 -credit courses and received two As, three Bs, four Cs, and one \(D\). a. Compute this student's grade-point average. b. Let the random variable \(X\) denote the number of points corresponding to a given letter grade. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Both the variance and the standard deviation of a random variable measure the spread of a probability distribution.

A panel of 64 economists was asked to predict the average unemployment rate for the upcoming year. The results of the survey follow: $$\begin{array}{lccccccc} \hline \text { Unemployment } & & & & & & & \\ \text { Rate, \% } & 4.5 & 4.6 & 4.7 & 4.8 & 4.9 & 5.0 & 5.1 \\ \hline \text { Economists } & 2 & 4 & 8 & 20 & 14 & 12 & 4 \\ \hline\end{array}$$ Based on this survey, what does the panel expect the average unemployment rate to be next year?

A panel of 50 economists was asked to predict the average prime interest rate for the upcoming year. The results of the survey follow: $$\begin{array}{lcccccc}\hline \text { Interest Rate, } \% & 4.9 & 5.0 & 5.1 & 5.2 & 5.3 & 5.4 \\\\\hline \text { Economists } & 3 & 8 & 12 & 14 &8 & 5 \\ \hline\end{array}$$ Based on this survey, what does the panel expect the average prime interest rate to be next year?

Paul Hunt is considering two business ventures. The anticipated returns (in thousands of dollars) of each venture are described by the following probability distributions: $$\begin{array}{l}\text { Venture } \mathrm{A}\\\\\begin{array}{cc} \hline \text { Earnings } & \text { Probability } \\\\\hline-20 & .3 \\ \hline 40 & .4 \\\\\hline 50 & .3 \\ \hline\end{array}\end{array}$$ $$\begin{array}{l}\text { Venture } \overline{\mathrm{B}}\\\\\begin{array}{cc} \hline \text { Earnings } & \text { Probability } \\\\\hline-15 & .2 \\ \hline 30 & .5 \\\\\hline 40 & 3 \\ \hline\end{array}\end{array}$$ a. Compute the mean and variance for each venture. b. Which investment would provide Paul with the higher expected return (the greater mean)? c. In which investment would the element of risk be less (that is, which probability distribution has the smaller variance)?
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