Question

this exercise is on probabilities and confidence intervals.
a. if two people are selected at random, the probability that they do not have the same birthday (day and month) is $\frac{365}{365}cdot\frac{364}{365}$. explain why this is so (ignore leap - years and assume 365 days in a year.)
the first person can have any birthday, so they can have a birthday on (square) of the 365 days. in order for the second person to not have the same birthday they must have one of the (square) remaining birthdays.
(type whole numbers.)
b. if three people are selected at random, find the probability that they all have different birthdays.
the probability that they all have different birthdays is (square)
(round to three decimal places as needed.)
c. if three people are selected at random, find the probability that at least two of them have the same birthday.
the probability that at least two of them have the same birthday is (square)
(round to three decimal places as needed.)
d. if 22 people are selected at random, find the probability that at least 2 of them have the same birthday.
the probability that at least two of them have the same birthday is (square)

Explanation:

Step1: Analyze two - person case

The first person can have a birthday on any of the 365 days. For the second person to have a different birthday, they must have one of the 364 remaining birthdays.

Step2: Calculate probability for three - different - birthdays case

For three people, the first person has 365 choices, the second has 364 choices, and the third has 363 choices. The probability $P(\text{all different})$ is $\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}=\frac{365\times364\times363}{365^{3}}\approx0.992$.

Step3: Calculate probability of at - least - two - same for three people

The probability that at least two of them have the same birthday is the complement of the event that all have different birthdays. So $P(\text{at least two same}) = 1 - P(\text{all different})=1 - 0.992 = 0.008$.

Step4: Calculate probability for 22 - people case

The probability that all 22 people have different birthdays is $P_{22}=\frac{365}{365}\times\frac{364}{365}\times\cdots\times\frac{365 - 21}{365}=\prod_{k = 0}^{21}\frac{365 - k}{365}\approx0.456$.
The probability that at least two of the 22 people have the same birthday is $1 - P_{22}=1 - 0.456 = 0.544$.

Answer:

a. The first person can have a birthday on 365 of the 365 days. In order for the second person to not have the same birthday they must have one of the 364 remaining birthdays.
b. 0.992
c. 0.008
d. 0.544