Suppose that six people work in the same office. Given that their birthdays were independen of each other, and the births are equally likely to occur in any month. Find the probability that
(a) None of these six people's birthdays fall in the same month.
(b) all six birthdays fall in the same month.

We are asked to compute two probabilities related to the distribution of birthdays among six people in the same office. The assumption is that each birthday is independent, and that any month is equally likely for a birthday (so 12 months in total).

### Part (a): Probability that none of the six people share the same birth month
This can be modeled as a classical **birthday problem** where we want all six birthdays to occur in different months. 

#### Approach:
- There are 12 months, and for the first person, there are 12 possible months they could be born in.
- For the second person, to avoid sharing a birthday with the first, they must be born in one of the remaining 11 months.
- For the third person, they must be born in one of the remaining 10 months, and so on.

Thus, the number of favorable outcomes (i.e., no shared months) is:

\[
12 \times 11 \times 10 \times 9 \times 8 \times 7
\]

The total number of possible outcomes, without any restriction (i.e., any of the six people could be born in any of the 12 months), is:

\[
12^6
\]

Thus, the probability that no one shares a birth month is:

\[
P(\text{none share a birth month}) = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{12^6}
\]

We can simplify this expression:

\[
P(\text{none share a birth month}) = \frac{12!}{(12-6)! \times 12^6}
= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{12^6}
\]

This can be computed directly.

### Part (b): Probability that all six birthdays fall in the same month
Here, the condition is that all six people have the same birth month.

#### Approach:
- For the first person, there are 12 possible months they could be born in.
- For the second person, they must be born in the same month as the first person, so there is only 1 option, and similarly for each subsequent person.

Thus, the number of favorable outcomes (i.e., all six are born in the same month) is:

\[
12 \times 1 \times 1 \times 1 \times 1 \times 1 = 12
\]

The total number of possible outcomes, as before, is:

\[
12^6
\]

Thus, the probability that all six people are born in the same month is:

\[
P(\text{all share the same birth month}) = \frac{12}{12^6} = \frac{1}{12^5}
\]

### Final answers:

(a) The probability that none of the six people share a birth month is:

\[
P(\text{none share a birth month}) = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{12^6}
\]

(b) The probability that all six people share the same birth month is:

\[
P(\text{all share the same birth month}) = \frac{1}{12^5}
\]

Would you like me to calculate these probabilities explicitly? Let me know if you have any questions!

Here are 5 related questions:
1. What is the probability that exactly two people share a birthday?
2. How does the probability change if we consider 10 people instead of 6?
3. What is the expected number of people who share a birthday in a group of 20?
4. How can this problem be extended to account for leap years?
5. What is the probability that at least two people share a birthday in a group of 6?

**Tip:** When solving probability problems, make sure to clearly distinguish between favorable outcomes and the total number of possible outcomes.

Suppose that six people work in the same office. Given that their birthdays were independent of each other, and the births are equally likely to occur in any month. Find the probability that (a) None of these six people's birthdays fall in the same month. (b) all six birthdays fall in the same month.

Explore the probability that none or all of six people in an office share the same birth month. Learn how combinatorics and probability principles are applied to solve these birthday problems, including the formula for finding the probability when no one shares a birth month and when all six share the same month.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation