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Chapter 15: Problem 7

In a club with 500 members, what is the probability that exactly two people have birthdays on July 4?

Short Answer

Expert verified
The probability that exactly two people have birthdays on July 4 in a club of 500 members is approximately 0.2401.

Step by step solution

01

Understand the Problem

Identify the key components for solving the problem. We need to find the probability that exactly two out of 500 people have their birthdays on July 4.
02

Identify the Distribution

Since we are dealing with a set number of trials (500) and each trial (each person's birthday) has two possible outcomes (having a birthday on July 4 or not), this is a binomial distribution problem.
03

Define Binomial Parameters

The parameters for binomial distribution are:\[n = 500\] (number of trials) and the probability of success (having a birthday on July 4) is \[p = \frac{1}{365}\].
04

Write the Binomial Probability Formula

The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(k = 2\) is the number of successes.
05

Substitute the Values

Substitute \(n = 500\), \(k = 2\), and \(p = \frac{1}{365}\) into the formula: \[ P(X = 2) = \binom{500}{2} \times \frac{1}{365^2} \times \bigg(1 - \frac{1}{365}\bigg)^{498} \]
06

Calculate the Combinations

Calculate \(\binom{500}{2}\): \[ \binom{500}{2} = \frac{500!}{2!(500-2)!} = \frac{500 \times 499}{2 \times 1} = 124750 \]
07

Compute the Probability

Calculate the entire expression: \[ P(X = 2) = 124750 \times \frac{1}{365^2} \times \bigg(\frac{364}{365}\bigg)^{498} \]
08

Simplify and Finalize the Result

After performing the calculations, the probability is approximately \( P(X = 2) \approx 0.2401 \)

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Key Concepts

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

Probability theory
Probability theory is a branch of mathematics concerned with analyzing random events. The base idea is to determine the likelihood of different outcomes. Probabilities range from 0 to 1, where 0 means an event will never occur, and 1 means it is certain to happen.

In our problem, we are dealing with a discrete probability distribution, which focuses on events happening in a set of distinct trials. Each trial represents one of the club members' birthdays.

Understanding the fundamentals of probability theory helps frame the problem and guides us in using appropriate formulas and strategies. This example of exact birthday matching is a classic application of these theoretical principles.
Binomial probability formula
The binomial probability formula calculates the probability of achieving a specific number of successes in a fixed number of independent trials. Each trial has a binary outcome: success (birth in our case) or failure (no birth).

The formula is written as:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

Here,
  • stands for the number of trials (500 members)
  • k is the desired number of successes (2 people sharing the same birthday)
  • p is the probability of success on a single trial ( \frac{1}{365}
  • \binom{n}{k} is the number of combinations representing different ways to choose k successes out of n trials.
.

This formula helps us calculate the probability of exactly 2 out of 500 people having their birthdays on July 4.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations of objects. It is essential in probability theory because it enumerates the ways that events can occur.

To solve our problem, we need to calculate the number of ways exactly 2 people out of 500 can share a birthday. This is given by the combination formula:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

For our problem,

  • n=500
  • k=2


The combination formula helps calculate:

\binom{500}{2} = \frac{500 x 499}{2 x 1} = 124750.

Combining these principles gives us a clear path to solving complex probability problems by using mathematical methods effectively.

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Most popular questions from this chapter

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Three coins are tossed; \(x=\) number of heads minus number of tails.

(a) Find the probability density function \(f(x)\) for the position \(x\) of a particle which is executing simple harmonic motion on \((-a, a)\) along the \(x\) axis. (See Chapter 7, Section 2, for a discussion of simple harmonic motion.) Hint: The value of \(x\) at time \(t\) is \(x=a\) cos \(\omega t .\) Find the velocity \(d x / d t ;\) then the probability of finding the particle in a given \(d x\) is proportional to the time it spends there which is inversely proportional to its speed there. Don't forget that the total probability of finding the particle somewhere must be 1. (b) Sketch the probability density function \(f(x)\) found in part (a) and also the cumulative distribution function \(F(x) \text { [see equation }(6.4)]\). (c) Find the average and the standard deviation of \(x\) in part (a).

Five cards are dealt from a shuffled deck. What is the probability that they are all of the same suit? That they are all diamond? That they are all face cards? That the five cards are a sequence in the same suit (for example, 3,4,5,6,7 of hearts)?

What is the probability that a number \(n, 1 \leq n \leq 99,\) is divisible by both 6 and \(10 ?\) By either 6 or 10 or both?

Consider the set of all permutations of the numbers \(1,2,3 .\) If you select a permutation at random, what is the probability that the number 2 is in the middle position? In the first position? Do your answers suggest a simple way of answering the same questions for the set of all permutations of the numbers 1 to \(7 ?\)
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