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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

Two cards are drawn at random from a shuffled deck and laid aside without being examined. Then a third card is drawn. Show that the probability that the third card is a spade is \(\frac{1}{4}\) just as it was for the first card. Hint: Consider all the (mutually exclusive) possibilities (two discarded cards spades, third card spade or not spade, etc.).

A card is selected from a shuffled deck. What is the probability that it is either a king or a club? That it is both a king and a club?

Given that a particle is inside a sphere of radius \(1,\) and that it has equal probabilities of being found in any two volume elements of the same size, find the cumulative distribution function \(F(r)\) for the spherical coordinate \(r,\) and from it find the density function \(f(r) .\) Hint: \(F(r)\) is the probability that the particle is inside a sphere of radius \(r .\) Find \(\overline{r}\) and \(\sigma\).

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) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let \(x=\) number of 3 's showing. Set up the sample space and associated probabilities for \(x.\) (b) What is the probability of at least one \(3 ?\) (c) If at least one 3 turns up, what is the probability that you picked the first box? (d) Find \(\bar{x}\) and \(\sigma\)

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