/*! 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 41 There are 12 signs of the Zodiac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There are 12 signs of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagittarius, Capricom, 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?

Short Answer

Expert verified
The probability of at least two people in the group of five having the same Zodiac sign is approximately 0.97 or 97%.

Step by step solution

01

Calculate the number of ways to have different signs for all 5 people

Let's first find the number of ways to assign each of the 5 people to a different Zodiac sign so that none of them share the same sign. The first person can be assigned any of the 12 signs, while the second person can be assigned any of the remaining 11 signs. Similarly, the third person can be assigned any of the remaining 10 signs, and so on. So, the total number of possible ways to assign different signs to all 5 people can be calculated as follows: Number of ways = 12 × 11 × 10 × 9 × 8
02

Calculate the total number of ways to assign signs to different people

We know that each person can belong to any of the 12 signs. So, for 5 people, the total number of ways to assign signs to different people is: Total number of ways = 12^5
03

Calculate the probability of all different signs

Now that we have both the number of ways for all different signs and the total number of ways to assign signs, let's calculate the probability that all 5 people have different signs: Probability of all different signs = \(\frac{\text{Number of ways with different signs}}{\text{Total number of ways}}\) = \(\frac{12 \times 11 \times 10 \times 9 \times 8}{12^5}\)
04

Calculate the probability of at least two of them having the same sign

To calculate the probability that at least two people have the same sign in the group, we will subtract the probability of all of them having different signs from 1: Probability of at least two people having the same sign = 1 - Probability of all different signs = \(1 - \frac{12 \times 11 \times 10 \times 9 \times 8}{12^5}\) Now, let's calculate this probability: Probability of at least two people having the same sign ≈ 0.97 So, the probability of at least two people in the group of five having the same Zodiac sign is approximately 0.97 or 97%.

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.

Zodiac Signs
The zodiac signs are a popular element of astrology, representing 12 distinct periods of the year. Each sign lasts approximately one month and is associated with specific dates in the calendar, often starting around the 20th of one month and ending around the 20th of the next. This cycle corresponds to the progression of the sun through a series of constellations as observed from Earth. The concept of zodiac signs dates back to ancient civilizations, who used the stars for navigation and the calendar for agricultural activities. In astrology, each sign is believed to bestow particular characteristics and traits on individuals born under it.
While the scientific basis of personality traits related to zodiac signs is debated, these signs are widely recognized and referenced in cultural contexts, such as horoscopes. Some of the common names for zodiac signs include Aries associated with adventurous traits, Libra with diplomacy, and Scorpio with intensity. The interest in zodiac signs often relates to the belief in cosmic influence on personal lives.
Combinatorial Analysis
Combinatorial analysis is a branch of mathematics focusing on counting, arranging, and analyzing objects. It's used to solve problems related to arrangements and selections under specific rules or constraints. When determining how zodiac signs can be distributed among a group of people, combinatorial analysis helps us calculate the number of possible arrangements.
In our problem, we use combinatorial analysis to find out how five people can have all different zodiac signs. We start by finding the number of ways five people can be independently assigned different signs out of the twelve available. This uses sequential multiplication, starting with 12 options for the first person, leaving 11 for the second, and continuing this way. This gives us a total of 12 × 11 × 10 × 9 × 8 possible ways to assign different signs to all five people, ensuring no repetition.
By understanding combinatorial principles, one can determine outcomes in more complex scenarios, such as determining the arrangements of various items where some may be indistinguishable from others or follow specific rules.
Probability Calculation
Probability calculation involves determining the likelihood of one or more events occurring. It's a fundamental aspect of probability theory and vital in assessing potential outcomes in uncertain situations. To solve the given problem, we are tasked with finding the probability of at least two people in a group of five having the same zodiac sign.
We start by identifying the probability that each of the five people has a different zodiac sign. This ratio compares the number of favorable outcomes (different signs) to the total possible outcomes for sign assignments. Using our previous combinatorial calculation, we obtain the probability:
  • Probability of all different signs = \( \frac{12 \times 11 \times 10 \times 9 \times 8}{12^5} \)
To find the reverse probability (at least two sharing a sign), we use the complement rule. The probability of at least two individuals sharing the same zodiac sign is:
  • 1 minus the probability of all different signs
Thus, we find the final probability to be approximately 0.97, or 97%, indicating a high likelihood of shared birthdays. This method illuminates how complementary counting is valuable in probability, making it easier to calculate complex probabilities by focusing on simpler opposite conditions first.

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

If a 5-card poker hand is dealt from a well-shuffled deck of 52 cards, what is the probability of being dealt the given hand? A flush (but not a straight flush)

According to a study conducted in 2003 concerning the participation, by age, of \(401(\mathrm{k})\) investors, the following data were obtained: $$ \begin{array}{lccccc} \hline \text { Age } & 20 \mathrm{~s} & 30 \mathrm{~s} & 40 \mathrm{~s} & 50 \mathrm{~s} & 60 \mathrm{~s} \\ \hline \text { Percent } & 11 & 28 & 32 & 22 & 7 \\ \hline \end{array} $$ a. What is the probability that a \(401(\mathrm{k})\) investor selected at random is in his or her 20 s or 60 s? b. What is the probability that a \(401(\mathrm{k})\) investor selected at random is under the age of 50 ?

Fifty people are selected at random. What is the probability that none of the people in this group have the same birthday?

CAR THEFT Figures obtained from a city's police department seem to indicate that, of all motor vehicles reported as stolen, \(64 \%\) were stolen by professionals whereas \(36 \%\) were stolen by amateurs (primarily for joy rides). Of those vehicles presumed stolen by professionals, \(24 \%\) were recovered within \(48 \mathrm{hr}, 16 \%\) were recovered after \(48 \mathrm{hr}\), and \(60 \%\) were never recovered. Of those vehicles presumed stolen by amateurs, \(38 \%\) were recovered within \(48 \mathrm{hr}, 58 \%\) were recovered after \(48 \mathrm{hr}\), and \(4 \%\) were never recovered. a. Draw a tree diagram representing these data. b. What is the probability that a vehicle stolen by a professional in this city will be recovered within \(48 \mathrm{hr}\) ? c. What is the probability that a vehicle stolen in this city will never be recovered?

An insurance company has compiled the accompanying data relating the age of drivers and the accident rate (the probability of being involved in an accident during a 1-yr period) for drivers within that group: $$\begin{array}{lcc} \hline & \begin{array}{c} \text { Percent of } \\ \text { Insured Drivers } \end{array} & \begin{array}{c} \text { Accident } \\ \text { Rate, \% } \end{array} \\ \hline \text { Under 25 } & 16 & 5.5 \\ \hline 25-44 & 40 & 2.5 \\ \hline 45-64 & 30 & 2 \\ \hline 65 \text { and over } & 14 & 4 \\ \hline \end{array}$$ What is the probability that an insured driver selected at random a. Will be involved in an accident during a particular 1 -yr period? b. Who is involved in an accident is under 25 ?

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.