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Determine the probability that at least 2 people in a room of 10 people share the same birthday, ignoring leap years and assuming each birthday is equally likely by answering the following questions: (a) Compute the probability that 10 people have different birthdays. (Hint: The first person's birthday can occur 365 ways; the second person's birthday can occur 364 ways, because he or she cannot have the same birthday as the first person; the third person's birthday can occur 363 ways, because he or she cannot have the same birthday as the first or second person; and so on.) (b) The complement of "10 people have different birthdays" is "at least 2 share a birthday." Use this information to compute the probability that at least 2 people out of 10 share the same birthday.

Step by step solution

01

Understand the problem

To find the probability that at least 2 people in a room of 10 people share the same birthday, we will first compute the probability that all 10 people have different birthdays and then use the complement rule.

02

Compute the number of possible outcomes

Each person can have a birthday on any of the 365 days. So, the total number of possible outcomes for 10 people is \(365^{10}\).

03

Compute the number of favorable outcomes

For the first person, there are 365 choices. For the second person, there are 364 choices (since they cannot have the same birthday as the first person). Continue this process for all 10 people: ewline \(365 \times 364 \times 363 \times 362 \times 361 \times 360 \times 359 \times 358 \times 357 \times 356\)

04

Divide the favorable outcomes by the total outcomes

The probability that all 10 people have different birthdays is obtained by dividing the number of favorable outcomes by the total outcomes: ewline \( \frac{365 \times 364 \times 363 \times 362 \times 361 \times 360 \times 359 \times 358 \times 357 \times 356}{365^{10}}\)

05

Simplify the expression

Simplify the expression to make it easier to calculate: ewline \( \frac{365!}{(365-10)!} \times \frac{1}{365^{10}} \)

06

Calculate the probability

Use a calculator to find the value: ewline \( \frac{365!}{(365-10)!} \times \frac{1}{365^{10}} \approx 0.883\)

07

Apply the complement rule

The complement of the event '10 people have different birthdays' is 'at least 2 share a birthday'. Thus, the probability that at least 2 people share a birthday is: ewline \(1 - 0.883 = 0.117\)

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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 famous problem in probability theory, which often surprises people with its counterintuitive result. It delves into the probability that in a group of just 23 people, at least two of them share the same birthday. The term 'paradox' stems from the seemingly low number of people needed for a high chance of shared birthdays.
The paradox relies on calculating the probability of no shared birthdays first, then using the complement rule to find the probability of at least one shared birthday. By considering that each person's birthday is equally likely to fall on any of the 365 days of the year (ignoring leap years), the calculations make use of permutations to determine possible outcomes.

complement rule

In probability theory, the complement rule is a powerful tool. It states that the probability of an event happening is equal to 1 minus the probability of the event not happening. This can be summarized as:
\[ P(A') = 1 - P(A) \]
In the context of the birthday paradox, the event A is 'at least two people share a birthday.' To find its probability, we first compute the probability of event A': 'no one shares a birthday.' We calculate the probability that all 10 people have unique birthdays and then use the complement rule:
\[ P(A) = 1 - P(A') \]
This approach simplifies complex probability problems and ensures a correct solution by focusing on the simpler complementary event.

permutations

Permutations play a critical role in calculating probabilities involving unique assignments or arrangements, such as the birthday paradox. A permutation considers the arrangement of a set where order matters. When determining the probability that each of 10 individuals has a distinct birthday, we deal with permutations of 365 days taken 10 at a time. The calculation follows that:
For the first person: 365 choices, second person: 364 choices, and so on until 356 choices for the 10th person:
\[ 365 \times 364 \times 363 \times ... \times 356 \]
Dividing this by the total number of possible birthday combinations (365^10), we derive the probability of all different birthdays. Understanding permutations helps simplify and solve complex counting problems in probability.