/*! 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 19 In Exercises 18, 20, and 21 assu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 19

In Exercises 18, 20, and 21 assume that the year has 366 days and all birthdays are equally likely. In Exercise 19 assume it is equally likely that a person is born in any given month of the year. a) What is the probability that two people chosen at random were born during the same month of the year? b) What is the probability that in a group of n people chosen at random, there are at least two born in the same month of the year? c) How many people chosen at random are needed to make the probability greater than 1?2 that there are at least two people born in the same month of the year?

Short Answer

Expert verified
a) \(\frac{1}{12}\), b) \(1 - \frac{12 \times 11 \cdots (12-n+1)}{12^n}\), c) n = 5.

Step by step solution

01

- Understanding the problem

For this exercise, we need to calculate different probabilities related to people being born in the same month. The year is assumed to have 12 months.
02

- Calculate the probability for part (a)

The total possible ways both individuals can be born in any month is 12. The number of favorable outcomes where both are born in the same month is also 12 (one for each month). Hence, the probability is \(\frac{12}{12^2} = \frac{1}{12}\).
03

- Calculate the probability for part (b)

To find the probability that at least two out of n people share the same birth month: 1. Calculate the probability that all n people are born in different months. 2. Subtract this probability from 1.Assuming the first person can be born in any of the 12 months. The second in any of the remaining 11 months and so on. This gives: \( P(\text{all different}) = \frac{12}{12} \times \frac{11}{12} \times \frac{10}{12} \cdots \times \frac{12-n+1}{12} \). The probability that at least two people are born in the same month is then: \( 1 - P(\text{all different}) \).
04

- Calculate the number of people for part (c)

We need to find the smallest n for which the probability of at least two people sharing the same birth month is greater than \( \frac{1}{2} \).Using the formula from Step 3, set up the equation: \( 1 - P(\text{all different}) > \frac{1}{2} \).Calculate the smallest n that satisfies this condition. As the calculations can be lengthy, it can be simplistically solved or checked using approximation or logarithms to get n.

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.

birthday problem
The birthday problem is an intriguing probability puzzle that explores the likelihood of two people sharing the same birthday in a group. Despite the seemingly low probability, it reveals counterintuitive results. In our case, we deal with months instead of days. We need to determine the probability that, in a group of people, at least two individuals share the same birth month. This exercise helps highlight the surprising outcomes that probability can yield.
probability calculation
Probability calculation is central to solving the birthday problem. It's about determining the chances of a specific event happening. For part (a) of our exercise, the probability that two people share the same birth month is straightforward. We have 12 different months. If two people are chosen at random, the possible ways they can be born in any of the 12 months is 12 x 12 = 144. Since we want both individuals to be born in the same month, there are only 12 favorable outcomes. Thus, the probability is calculated as \(\frac{12}{144} = \frac{1}{12}\). This basic calculation sets the stage for more complex scenarios.
combinatorial probability
Combinatorial probability deals with calculating the likelihood of different combinations of events. For part (b) of the exercise, we use combinatorics to find the probability that at least two people share the same birth month in a group of \(n\) people. The approach involves:
  • Calculating the probability that all \(n\) people are born in different months.

  • Using the formula: \[ P(\text{all different}) = \frac{12}{12} \times \frac{11}{12} \times \frac{10}{12} \times \text{...} \times \frac{12-n+1}{12} \]

  • Finding the complementary probability that at least two people share the same month: \[ 1 - P(\text{all different}) \]
This method employs a step-by-step approach to combinatorial problems, helping us navigate through more complex probability scenarios.
statistical independence
Statistical independence means the occurrence of one event doesn't affect another. This principle is essential in our calculations. When calculating the probability in part (b), we assume the months chosen by each person are independent events. If one person is born in a particular month, it doesn't influence another person's birth month. This independence simplifies our probability calculations, allowing us to multiply probabilities step by step.
Statistical independence plays a crucial role in the birthday problem and many other probability-based scenarios. By understanding it, we can better analyze and solve complex problems involving multiple random events.

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

Use the law of total expectation to find the average weight of a breeding elephant seal, given that 12% of the breeding elephant seals are male and the rest are female, and the expected weights of a breeding elephant seal is 4200 pounds for a male and 1100 pounds for a female.

Let \(X\) be the number appearing on the first die when two fair dice are rolled and let \(Y\) be the sum of the numbers appearing on the two dice. Show that \(E(X) E(Y) \neq E(X Y) .\)

What is the probability that when a coin is flipped six times in a row, it lands heads up every time?

A pair of dice is loaded. The probability that a 4 appears on the first die is \(2 / 7,\) and the probability that a 3 appears on the second die is \(2 / 7 .\) Other outcomes for each die appear with probability \(1 / 7 .\) What is the probability of 7 appearing as the sum of the numbers when the two dice are rolled?

In this exercise we will use Bayes鈥� theorem to solve the Monty Hall puzzle (Example 10 in Section 7.1). Recall that in this puzzle you are asked to select one of three doors to open. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door; assume that p(W = k) = 1?3 for k = 1, 2, 3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i. a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors? b) Find p(M = j ? W = k) for j = 1, 2, 3 and k = 1, 2, 3. c) Use Bayes鈥� theorem to find p(W = j ? M = k) where i and j and k are distinct values. d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the chance to do so.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.