Problem 59 - Algebra and Trigonometry with Analytic Geometry

Question

Birthday probability (a) Show that the probability \(p\) that \(n\) people all have different birthdays is given by $$ p=\frac{365 !}{365^{n}(365-n) !} $$ (b) If a room contains 32 people, approximate the probability that two or more people have the same birthday. (First approximate \(\ln p\) by using the following formula from advanced mathematics: $$ \ln n ! \Leftrightarrow n \ln n-n .) $$

Step-by-Step Solution

Answer

The probability that two or more people have the same birthday among 32 people is nearly 1.

Step 1: Understand the Problem

We need to find the probability that in a group of \(n\) people, all have different birthdays. Then, using this, we'll find the probability that at least two people share a birthday in a group of 32 people.

Step 2: Formula for Different Birthdays

The probability that \(n\) people all have different birthdays is given by \(p = \frac{365 !}{365^{n}(365-n) !}\)

Step 3: Simplify Using Logs

To simplify calculations, we approximate \(\ln(n!)\) using Stirling's approximation: \(\ln(n!) \approx n \ln n - n\)

Step 4: Approximate \(\ln p\)

Using the formula for \(p\), we write:\(\ln p = \ln(365!) - n \ln(365) - \ln((365-n)!)\). We replace \(\ln(365!)\) and \(\ln((365-n)!)\) using Stirling's approximation:- \(\ln(365!) \approx 365 \ln 365 - 365\)- \(\ln((365-n)!) \approx (365-n) \ln(365-n) - (365-n)\)

Step 5: Substitute and Simplify

Substitute the approximations in the expression for \(\ln p\): \[ \ln p \approx (365 \ln 365 - 365) - n \ln 365 - ((365-n) \ln(365-n) - (365-n)) \] Simplify the expression: \[ \ln p \approx 365 \ln 365 - 365 - n \ln 365 - 365 \ln(365-n) + n + 365 - n \ln(365-n) \] \[ \ln p \approx 365 \ln 365 - n \ln 365 - (365-n) \ln(365-n) \]

Step 6: Numerical Calculation for 32 People

Substitute \(n = 32\): Calculate: - \(365 \ln 365 \approx 1975.833\)- \(32 \ln 365 \approx 1704.307\)- \(333 \ln 333 \approx 1761.505\) Therefore, \[ \ln p \approx 1975.833 - 1704.307 - 1761.505 = -1489.979\]

Step 7: Convert \(\ln p\) to Probability

Convert \(\ln p\) back to \(p\) using exponential:\(p = e^{-1489.979}\), which is extremely small, close to zero.

Step 8: Probability of Shared Birthday

The probability that two or more people share a birthday is\(1 - p\), which is very close to 1 since \(p\) is nearly zero.

Key Concepts

Understanding Probability Theory

Probability theory is a fundamental branch of mathematics focused on the analysis of random events. It helps us understand and quantify the likelihood of different outcomes. In simple terms, probability measures how likely an event is to happen. Values range from 0 to 1, where 0 means an event will not happen, and 1 means it will definitely happen.

In the context of the birthday problem, we are looking at the likelihood of all birthdays being different among a group of people. If everything is equally likely, the first person can choose any of the 365 days (ignoring leap years). The second person then has only 364 choices to avoid a same birthday, and so on. This sequence forms the basis of our probability calculation.

The formula for the probability, \( p = \frac{365!}{365^n(365-n)!} \), arises from these choices. The denominator \(365^n\) represents the total possible combinations of birthdays for \(n\) people. The numerator ensures that birthdays are all different."

• The closer \(p\) is to 0, the higher the chance that at least two people share a birthday.
• When we reach around 23 people, there's already more than a 50% chance of shared birthdays. This counterintuitive result showcases the non-absolute nature of probability.

Probability helps us identify and manage uncertainty in everyday and scientific contexts. It encourages evaluating not just outcomes, but their likelihood and how they interconnect.

Diving into Stirling's Approximation

Stirling's approximation is a powerful mathematical tool for approximating factorials, especially useful with large numbers. The factorial of a number \( n! \) grows extremely large as \( n \) increases, making calculations cumbersome.

Stirling's approximation simplifies this by providing an easier expression: \( \ln(n!) \approx n \ln n - n \). This helps us approximate calculations in probability and statistics more feasibly, such as in the birthday problem.

In our birthday probability, we used Stirling's approximation to manage the enormous factorials. Calculating \( 365! \) directly is impractical, but using Stirling's method, we simplify our logs, which then convert back into more manageable numerical values.

• This approximation doesn't give the exact value but simplifies the otherwise complex factorial calculations.
• It holds especially well for higher \( n \), making it ideal for large datasets or complex probabilities.

By incorporating this approach, we're able to handle and interpret probabilities that would otherwise require vast computational resources.

Exploring Permutations in Probability

Permutations are about arranging objects in specific orders, a cornerstone concept in counting probabilities. In probability theory, permutations help calculate possible arrangements of events, critical when order matters, like in the very structure of the birthday problem.

If we consider \( n \) people, the permutation involves how many different ways we can assign them different birthdays. This is embedded in the expression \( 365 \times 364 \times ... \text{(down to )} \times (365-n+1) \), representing the successive reduction in choices for each new individual without repetition.

Each of these decreasing numbers corresponds to a new entrant in the group being assigned an available birthday from the remaining pool. This method directly ties into calculating \( \frac{365!}{(365-n)!} \), which is the numerator of our main probability formula.

• Understanding permutations clarifies why the chance of unique birthdays decreases with more people.
• Incorporating permutations, we systematically assess each fresh probability impact.

Permutations not only show us arrangements but underpin the interdependencies of event assumptions critical in more complex statistical assessments.