Chapter 8: Problem 4
Consider the Birthday Problem, ignoring leap years. Determine the probability that two people in your class have the same birthday under each of the following circumstances: (a) There are 20 people in your class. (b) There are 30 people in your class.
Short Answer
Expert verified
(a) 0.41144; (b) 0.70632.
Step by step solution
01
Understand the Problem
The Birthday Problem asks us to find the probability that at least two people in a group have the same birthday. In this exercise, we will consider two different class sizes to find this probability.
02
Complementary Probability Approach
It is easier to calculate the probability that no two people have the same birthday and then subtract it from 1. This involves understanding the total number of ways to assign different birthdays and the number of ways to assign birthdays such that no one shares a birthday.
03
Calculate Probability for 20 People
For 20 people, the probability that no two people share a birthday is calculated as:\[P(20) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \cdots \times \frac{346}{365}\]This results in:\[P(20) \approx 0.58856\]Thus, the probability that at least two people have the same birthday is:\[1 - P(20) \approx 1 - 0.58856 = 0.41144\]
04
Calculate Probability for 30 People
For 30 people, the probability that no two people share a birthday is calculated as:\[P(30) = \frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{336}{365}\]This results in:\[P(30) \approx 0.29368\]Thus, the probability that at least two people have the same birthday is:\[1 - P(30) \approx 1 - 0.29368 = 0.70632\]
05
Conclude with Calculated Probabilities
We have calculated that the probability of at least two people having the same birthday is approximately 0.41144 for a class size of 20 people and 0.70632 for a class size of 30 people.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Birthday Problem
The Birthday Problem is a classic example in probability theory that surprises many with its counterintuitive results. The problem asks how likely it is for at least two people in a room to share the same birthday. Surprisingly, the probability is higher than what we might expect. This problem illustrates the principle of permutation and the sheer number of possible combinations that can occur in discrete mathematics. Although it ignores leap years, which simplifies the calculation, the result often leads to better intuition around random events and probability.
Complementary Probability
Complementary Probability is a useful approach in this problem because calculating the probability of at least two people sharing a birthday directly can be complex. Instead, it's easier to calculate the opposite—what's the probability that **no one** shares a birthday—and then subtract this value from 1. Generally, if the event is "A," the complementary event, "not A," allows us to use the relation:
\[ P(A) = 1 - P( ext{not A}) \]
For the Birthday Problem, "not A" would be everyone having a unique birthday, making the calculation manageable.
\[ P(A) = 1 - P( ext{not A}) \]
For the Birthday Problem, "not A" would be everyone having a unique birthday, making the calculation manageable.
Probability Calculation
Probability Calculation for the Birthday Problem involves understanding the permutations of different birthdays assigned to individuals. For example, with 20 people on 365 days, the first person can have any birthday, the second has 364 choices, and so on.
The formula to find the probability that no two people share a birthday when there are 20 people:
\[ P(20) = \frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{346}{365} \]
The result of this multiplication provides the probability that no one shares a birthday. For class sizes of 20 and 30, the exact computations reduce to a simple subtraction from 1 to find the probability of at least two people sharing a birthday.
The formula to find the probability that no two people share a birthday when there are 20 people:
\[ P(20) = \frac{365}{365} \times \frac{364}{365} \times \cdots \times \frac{346}{365} \]
The result of this multiplication provides the probability that no one shares a birthday. For class sizes of 20 and 30, the exact computations reduce to a simple subtraction from 1 to find the probability of at least two people sharing a birthday.
Discrete Mathematics
Discrete Mathematics deals with distinct and separate values or objects, which aligns perfectly with the Birthday Problem. This mathematical field involves problems with finite sets—like birthdays, where each person can only have one from a limited pool of 365 days. The Birthday Problem utilizes concepts like permutations and probability theory, crucial to understanding discrete systems. These concepts form the backbone of computer algorithms, cryptography, and data structures frequently used across sciences and engineering.