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Chapter 5: Problem 49

Show that with 25 students, there are 300 pairs of students who can have the same birthday. So it's really not so surprising if at least two students have the same birthday.

Short Answer

Expert verified
There are 300 pairs of students who can have the same birthday among 25 students, making it likely that at least two share a birthday.

Step by step solution

01

Recognize the Problem Context

We are given 25 students and we need to find the number of ways to pair them in such a way that at least two students can have the same birthday.
02

Understand Pair Counting

To find out how many pairs we can form from 25 students, we use combinations. Specifically, we use the formula for combinations to determine how many ways we can select 2 students from 25.
03

Apply Combinations Formula

Use the combinations formula: \( \binom{n}{k} \), where \(n\) is the total number of students (25), and \(k\) is the number of students in each pair (2). Calculate \( \binom{25}{2} \).
04

Calculate the Combinations Result

Calculate \( \binom{25}{2} = \frac{25!}{2!(25-2)!} = \frac{25 \times 24}{2 \times 1} = 300 \). This represents the total number of unique pairs.
05

Interpret the Result

The result of 300 pairs means there are 300 different possible ways for two students out of 25 to share the same birthday. This high number makes it quite likely that at least two students will indeed have the same birthday.

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Key Concepts

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

Birthday Paradox
The Birthday Paradox is a fascinating concept in probability theory. Despite being called a 'paradox', it is not actually a contradiction but rather a surprising reality about probabilities.
When you walk into a room with people, the likelihood that any two people share the same birthday isn't as low as you might expect.
The key observation here is that the probability calculation doesn't work the way we intuitively imagine the chances increase. Here's what's interesting:
  • You might think you need hundreds of people to have a good chance of a shared birthday.
  • Surprisingly, with just 23 people, there's already a 50% chance that two people will share a birthday.
  • With 25 people, the chance of sharing a birthday is already significantly above 50%.
This counterintuitive result is due to the large number of pairings possible in even small groups, vastly increasing the probability of a matching birthday.
Combination Formula
In combinatorics, the combination formula is a crucial tool. It's what helps us calculate the number of possible pairings or selections from a larger group when the order doesn't matter.
This is vital in determining the number of student pairs who could share a birthday.The formula is expressed as:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]Where:
  • \(n\) is the total number of items (e.g., students).
  • \(k\) is the number of items to choose.
  • \(!\) denotes factorial, which means you multiply it by all smaller positive integers.
In the given problem, we determined the number of ways to choose 2 students out of 25 using the combination formula: \(\binom{25}{2}\).
Calculating this gives us 300, meaning there are 300 different pairs of students who could share a birthday.
Probability Theory
Probability theory is an essential area of mathematics that helps us understand and quantify uncertainties. It's the foundation of many concepts, including the Birthday Paradox.
Probabilities are calculated as fractions or percentages that represent the likelihood of an event occurring. With probability theory, you learn that:
  • Events with high probability will likely occur.
  • Events with low probability are less likely but can still happen.
Determining shared birthdays among a group of people uses probability theory.
Given the large number of potential pairings in a group, these combinations lead to higher probabilities of shared birthdays than one might assume.
This makes probability theory incredibly useful for making sense of outcomes and making predictions based on numerical data and likely scenarios.

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Most popular questions from this chapter

Your friend decides to flip a coin repeatedly to analyze whether the probability of a head on each flip is \(1 / 2\). He flips the coin 10 times and observes a head 7 times. He concludes that the probability of a head for this coin is \(7 / 10=0.70 .\) a. Your friend claims that the coin is not balanced, since the probability is \(n o t 0.50\). What's wrong with your friend's claim? b. If the probability of flipping a head is actually \(1 / 2,\) what would you have to do to ensure that the cumulative proportion of heads falls very close to \(1 / 2 ?\)

Your teacher gives a truefalse pop quiz with 10 questions. a. Show that the number of possible outcomes for the sample space of possible sequences of 10 answers is 1024 . b. What is the complement of the event of getting at least one of the questions wrong? c. With random guessing, show that the probability of getting at least one question wrong is \(0.999 .\)

A survey asks subjects whether they believe that global warming is happening (yes or no) and how much fuel they plan to use annually for automobile driving in the future, compared to their past use (less, about the same, more). a. Show the sample space of possible outcomes by drawing a tree diagram that first gives the response on global warming and then the response on fuel use. b. Let \(A\) be the event of a "yes" response on global warming and let \(\mathrm{B}\) be the event of a "less" response on future fuel use. Suppose \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{B})>\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) .\) Indicate whether \(\mathrm{A}\) and \(\mathrm{B}\) are independent events, and explain what this means in nontechnical terms.

Two friends decide to go to the track and place some bets. One friend remarks that in an upcoming race, the number 5 horse is paying 50 to \(1 .\) This means that anyone who bets on the 5 horse receives \(\$ 50\) for each \(\$ 1\) bet, if in fact the 5 horse wins the race. He goes on to mention that it is a great bet, because there are only eight horses running in the race, and therefore the probability of horse 5 winning must be \(1 / 8 .\) Is the last statement true or false? Explain.

Workers specified as actively disengaged are those who are emotionally disconnected from their work and workplace. A Gallup poll conducted in December \(2010^{5}\) surveyed individuals who were either unemployed or who were actively disengaged in their current position. Individuals were asked to classify themselves as thriving or struggling. The poll reported that \(42 \%\) of the actively disengaged group claimed to be thriving, compared to \(48 \%\) of the unemployed group. a. Are these percentages (probabilities) ordinary or conditional? Explain, by specifying events to which the probabilities refer. b. Of the individuals polled, 1266 were unemployed and 400 were actively disengaged. Create a contingency table showing counts for job status and self- classification. c. Create a tree diagram such that the first branching represents job status and the second branching represent self-classification. Be sure to include the appropriate percentages on each branch.
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