Question

(round to three decimal places as needed.)
show that if 23 people are selected at random, the probability that at least 2 of them have the same birthday is greater than $\frac{1}{2}$.
the probability that none of the 23 people share a birthday is . the probability that at least 2 of 23 people have the same birthday is . the probability that there exists with the same birthday is than the probability of all 23 not sharing a birthday.
(round to three decimal places as needed.)

Explanation:

Step1: Calculate probability of no - shared birthdays

The first person can have a birthday on any of 365 days. The second person must have a birthday on one of the remaining 364 days, the third person on one of the remaining 363 days, and so on for 23 people. The probability $P(\text{no shared})$ that none of the 23 people share a birthday is given by the formula:
\[P(\text{no shared})=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\cdots\times\frac{365 - 22}{365}=\prod_{k = 0}^{22}\frac{365 - k}{365}\]
\[P(\text{no shared})=\frac{365!}{365^{23}(365 - 23)!}\approx0.493\]

Step2: Calculate probability of at least 2 people sharing birthdays

The probability $P(\text{at least 2 share})$ that at least 2 of the 23 people have the same birthday is the complement of the event that none of them share a birthday. So, $P(\text{at least 2 share})=1 - P(\text{no shared})$.
\[P(\text{at least 2 share})=1-0.493 = 0.507\]

Answer:

The probability that none of the 23 people share a birthday is approximately $0.493$. The probability that at least 2 of 23 people have the same birthday is approximately $0.507$. The probability that there exists 2 or more people with the same birthday is greater than the probability of all 23 not sharing a birthday.