/*! 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 Determine the probability that a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 33

Determine the probability that at least 2 people in a room of 10 people share the same birthday, ignoring leap years and assuming each birthday is equally likely, by answering the following questions: (a) Compute the probability that 10 people have 10 different birthdays. Hint: The first person's birthday can occur 365 ways, the second person's birthday can occur 364 ways, because he or she cannot have the same birthday as the first person, the third person's birthday can occur 363 ways, because he or she cannot have the same birthday as the first or second person, and so on. (b) The complement of "10 people have different birthdays" is "at least 2 share a birthday." Use this information to compute the probability that at least 2 people out of 10 share the same birthday.

Short Answer

Expert verified
The probability that at least 2 people out of 10 share the same birthday is \[ 1 - \frac{365 \times 364 \times 363 \times ... \times 356}{365^{10}} \].

Step by step solution

01

Calculate the total number of possible birthday combinations

For the first person, there are 365 possible birthdays. For each subsequent person, the number of possible birthdays remains 365 (since each birthday is equally likely).Thus, the total number of possible birthday combinations for 10 people is computed as: \[ 365^{10} \]
02

Calculate the number of ways 10 people can have different birthdays

For 10 people to all have different birthdays, the first person can have any of 365 birthdays, the second person can have any of the remaining 364 birthdays, the third person can have any of the 363 remaining birthdays, and so on. Therefore, the number of ways in which 10 people can have different birthdays is given by: \[ 365 \times 364 \times 363 \times ... \times 356 \]
03

Compute the probability that 10 people have different birthdays

The probability that all 10 people have different birthdays is given by the ratio of the number of favorable outcomes (10 people having different birthdays) to the total number of possible outcomes (any combination of birthdays). Therefore, \[ P(\text{different birthdays}) = \frac{365 \times 364 \times 363 \times ... \times 356}{365^{10}} \]
04

Compute the complement probability

The complement of the event that all 10 people have different birthdays is that at least two people share the same birthday. To find this probability, we subtract the probability of all different birthdays from 1: \[ P(\text{at least two share a birthday}) = 1 - P(\text{different birthdays}) \]

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 91影视!

Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur. It ranges from 0 to 1, where 0 means the event cannot happen and 1 means it will certainly happen. In the context of the birthday problem, we calculate the probability of certain events related to people's birthdays. When considering multiple possibilities, the total probability must equal 1. If you know the probability of one event, you can find the probability of its complement (the opposite event) by subtracting from 1. This concept is key in solving the birthday problem, where we look at the chance that at least two people share the same birthday by considering the opposite (that no one shares a birthday) first.
Combinatorics
Combinatorics involves counting, arranging, and finding patterns in sets of objects. For the birthday problem, we use combinatorial methods to count the number of ways people can have different or the same birthdays. We start by calculating the total number of possible birthday combinations. For 10 people, each having a birthday from 365 days, we have: \[ 365^{10} \]
Next, we determine the number of favorable outcomes for 10 people having different birthdays. This is done by multiplying the number of available days reducing by one each time: \[ 365 \times 364 \times 363 \times ... \times 356 \]
Understanding these basic principles of combinatorics helps us break down complex probability problems into simpler, countable units.
Birthday Problem
The birthday problem explores the counter-intuitive probability that in a group of people, at least two individuals will share the same birthday. Even with small groups, this probability can be surprisingly high.
For our problem with 10 people, we first compute the probability that no one shares a birthday by comparing the number of ways to have different birthdays to the total possible combinations. The fraction gives us the probability of different birthdays: \[ P(\text{different birthdays}) = \frac{365 \times 364 \times 363 \times ... \times 356}{365^{10}} \]
Finally, to find the chance that at least two people share a birthday, we subtract this probability from 1: \[ P(\text{at least two share a birthday}) = 1 - P(\text{different birthdays}) \]
This concept highlights the interesting results that probability and combinatorics can reveal about common experiences such as shared birthdays.

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

The following data represent the number of drivers involved in fatal crashes in the United States in 2013 by day of the week and gender. $$ \begin{array}{lrrr} & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Sunday } & 4143 & 2287 & 6430 \\ \hline \text { Monday } & 3178 & 1705 & 4883 \\ \hline \text { Tuesday } & 3280 & 1739 & 5019 \\ \hline \text { Wednesday } & 3197 & 1729 & 4926 \\ \hline \text { Thursday } & 3389 & 1839 & \mathbf{5 2 2 8} \\ \hline \text { Friday } & 3975 & 2179 & \mathbf{6 1 5 4} \\ \hline \text { Saturday } & 4749 & 2511 & \mathbf{7 2 6 0} \\ \hline \text { Total } & \mathbf{2 5 , 9 1 1} & \mathbf{1 3 , 9 8 9} & \mathbf{3 9 , 9 0 0} \\ \hline \end{array} $$ (a) Among Sunday fatal crashes, what is the probability that a randomly selected fatality is female? (b) Among female fatalities, what is the probability that a randomly selected fatality occurs on Sunday? (c) Are there any days in which a fatality is more likely to be male? That is, is \(P(\) male \(\mid\) Sunday \()\) much different from \(P(\) male \(\mid\) Monday \()\) and so on?

Suppose that a digital music player has 13 tracks. After listening to all the songs, you decide that you like 5 of them. With the random feature on your player, each of the 13 songs is played once in random order. Find the probability that among the first two songs played (a) You like both of them. Would this be unusual? (b) You like neither of them. (c) You like exactly one of them. (d) Redo (a)-(c) if a song can be replayed before all 13 songs are played (if, for example, track 2 can play twice in a row).

According to the U.S. Census Bureau, the probability that a randomly selected household speaks only English at home is \(0.81 .\) The probability that a randomly selected household speaks only Spanish at home is 0.12 . (a) What is the probability that a randomly selected household speaks only English or only Spanish at home? (b) What is the probability that a randomly selected household speaks a language other than only English or only Spanish at home? (c) What is the probability that a randomly selected household speaks a language other than only English at home? (d) Can the probability that a randomly selected household speaks only Polish at home equal 0.08 ? Why or why not?

According to the Statistical Abstract of the United States, about \(17 \%\) of all 18 - to 25 -year-olds are current marijuana users. (a) What is the probability that four randomly selected 18 - to 25-year-olds are all marijuana users? (b) What is the probability that among four randomly selected 18- to 25-year- olds at least one is a marijuana user?

A committee consisting of four women and three men will randomly select two people to attend a conference in Hawaii. Find the probability that both are women.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
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.