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Chapter 7: Problem 27

In a group of 2,000 people, must at least 5 have the same birthday? Why?

Short Answer

Expert verified
By applying the Pigeonhole Principle to a group of 2,000 people and 365 potential birthdays, we calculate that there must be at least \(\lceil \frac{2000}{365} \rceil \approx 5.48\) rounded up to 6 people with the same birthday. Therefore, yes, at least 5 people in the group must have the same birthday, and in fact, there should be at least 6 people with the same birthday.

Step by step solution

01

Apply Pigeonhole Principle to the problem.

As we mentioned earlier, the Pigeonhole Principle states that when there are more items (n) than categories (m), there will be at least one category with more than one item. In this problem, we'll treat each day of the year as a category (with 365 categories) and the 2,000 people as items.
02

Divide the number of people by the number of days in a year.

To apply this principle, we'll first need to find out how many times the number of people in the group exceeds the number of days in a year. We will do this by dividing the total number of people by the total number of days in a year (assuming it's not a leap year): \(P = \frac{n}{m}\) with: P = minimum number of people with the same birthday n = number of people (2,000) m = number of categories (365 days)
03

Calculate the value of P.

Now we'll substitute our values for n and m into the equation: \(P = \frac{2000}{365}\) \(P \approx 5.48\)
04

Round up the value of P and claim the result.

Since the Pigeonhole Principle is interested in whole items or people, we must round up the value we got in step 3 to the nearest whole number. In this case, we round up 5.48 to 6. This means that in this group of 2,000 people, there must be at least 6 people with the same birthday. So, the answer to the question is: Yes, at least 5 people must have the same birthday in this group. In fact, there should be at least 6 people with the same birthday.

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Key Concepts

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

Discrete Mathematics
Discrete Mathematics is a fundamental area of mathematics focused on structures that are fundamentally distinct and separate. It's all about countable, often finite, mathematical structures rather than continuous ones. In this field, you'll dive deep into concepts like sets, graphs, and algorithms.

One of the significant ideas in discrete mathematics is the Pigeonhole Principle. This simple yet powerful principle states that if you try to distribute more items than there are containers, at least one container must contain more than one item. It's a rather straightforward way to explain how limited resources or spaces are distributed among a larger group. For example, if you have ten pigeons and nine pigeonholes, one pigeonhole is going to end up with at least two pigeons.

Discrete mathematics is crucial because it forms the backbone of computer science and enables the study of algorithms, and data structures. It's like learning the grammar behind how computation and data processing work.
Birthday Problem
The birthday problem is a classic example of how seemingly intuitive probability can lead to surprising results. It involves determining the likelihood that, in a set of randomly chosen people, some share the same birthday. Its relevance is tied to the Pigeonhole Principle, as you will see.

When we talk about birthdays, there are 365 possible days each person could be born (ignoring leap years). If we bring together 23 people, six times less than 2000, which is our current exercise's scenario, there is more than a 50% chance that two of them share a birthday.

The mathematics behind this problem depends on calculating probabilities in reverse. Instead of trying to find the probability of a shared birthday occurring, we determine the probability of everyone's birthday falling on different days. When this probability is subtracted from 1, we are left with the probability of at least one shared birthday, showing how quickly the likelihood of shared birthdays grows.
Probability Theory
Probability Theory is the branch of mathematics concerned with analyzing random events. The goal is to express the likelihood of different outcomes using numbers between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

When considering the birthday problem, probability theory helps quantify the chances that at least two people in a group share the same birthday. With larger groups, like our group of 2,000 people, probability theory and the Pigeonhole Principle affirm that shared birthdays are not only possible but highly likely.

Probability theory is essential for understanding how events and data behave over time. It is integral to fields such as statistics, finance, science, and computer algorithms. By mastering these concepts, you can predict patterns and tendencies based on existing information.

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Most popular questions from this chapter

Prove that a union of any two countably infinite sets is countably infinite.

Let \(S\) be a set of ten integers chosen from 1 through 50 . Show that the set contains at least two different (but not necessarily disjoint) subsets of four integers that add up to the same number. (For instance, if the ten numbers are \(\\{3,8,9,18,24,34,35,41,44,50\\}\), the subsets can be taken to be \(\\{8,24,34,35\\}\) and \(\\{9,18,24,50\\}\). The numbers in both of these add up to 101.)

a. Define \(f: \mathbf{Z} \rightarrow \mathbf{Z}\) by the rule \(f(n)=2 n\), for all integers \(n\). (i) Is \(f\) one-to-one? Prove or give a counterexample. (ii) Is \(f\) onto? Prove or give a counterexample. b. Let \(2 \mathbf{Z}\) denote the set of all even integers. That is, \(2 \mathbf{Z}=\) \(\\{n \in \mathbf{Z} \mid n=2 k\), for some integer \(k\\}\). Define \(h: \mathbf{Z} \rightarrow 2 \mathbf{Z}\) by the rule \(h(n)=2 n\), for all integers \(n\). Is \(h\) onto? Prove or give a counterexample.

What is the largest number of elements that a set of integers from 1 through 100 can have so that no one element in the set is divisible by another? (Hint: Imagine writing all the numbers from 1 through 100 in the form \(2^{k} \cdot m\), where \(k \geq 0\) and \(m\) is odd.)

Prove that \(\mathbf{Z} \times \mathbf{Z}\), the Cartesian product of the set of integers with itself, is countably infinite.
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