Question

if 15 people are selected at random, find the probability that at least 2 of them have the same birthday. (round to three decimal places as needed.)

Explanation:

Step1: Find the probability of all - different birthdays

The first person can have a birthday on any of the 365 days of the year. The second person must have a birthday on a different day than the first, so the probability that the second person has a different birthday from the first is $\frac{364}{365}$. The third person must have a birthday different from the first two, so the probability that the third person has a different birthday from the first two is $\frac{363}{365}$, and so on. For $n = 15$ people, the probability that all 15 people have different birthdays is $P(\text{all different})=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\cdots\times\frac{365 - 14}{365}=\prod_{k = 0}^{14}\frac{365 - k}{365}$.

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LATEXBLOCK0
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Step2: Find the probability of at - least two people having the same birthday

The probability that at least two people have the same birthday is the complement of the event that all people have different birthdays. Let $P(X\geq2)$ be the probability that at least two people have the same birthday and $P(X = 0)$ be the probability that all people have different birthdays. Then $P(X\geq2)=1 - P(X = 0)$.
Since $P(X = 0)\approx0.747$, then $P(X\geq2)=1 - 0.747 = 0.253$.

Answer:

$0.253$