Question

e. 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 \\(square\\). the probability that at least 2 of 23 people have the same birthday is \\(square\\). the probability that there exists people with the same birthday is \\(square\\) than the probability of all 23 not sharing a birthday.
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the probability that none share a birthday

The formula for the probability that \(n\) people have different birthdays is \(P(\text{no match})=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\cdots\times\frac{365 - n+ 1}{365}\).
For \(n = 23\), \(P(\text{no match})=\frac{365!}{365^{23}(365 - 23)!}\)

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Step2: Calculate the probability that at least 2 share a birthday

Use the complement rule \(P(\text{at least one match})=1 - P(\text{no match})\)
Since \(P(\text{no match})\approx0.493\), then \(P(\text{at least one match})=1-0.493 = 0.507\)

Answer:

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