91Ó°ÊÓ

The given data is that at least 2 people have the same birthday in a group of 12 people.

There are 365 possibilities for each person's birthday. Since there are 12 people in the group, there are ${365}^{12}$possibilities for the birthday.

Thus, the number of elements in the sample space are $n\left(S\right)={365}^{12}$

let E be the event that at least two people have same birthday and $\overline{E}$be the event no two people have the same birthday.

First choose one person at random. There are 365 possibilities for his or her birthday. Next, choose the second person. If no two people have the same birthday then there are 364 possibilities for his or her birthday. Similarly, for 12th person, there are 354 possibilities for his birthday.

Thus,$n\left(E\right)=365·364·363·....354$

The probability of event $\overline{E}$is given by

$$\begin{gathered} P\left(\overline{E}\right)=\frac{n\left(\overline{E}\right)}{n\left(S\right)} \\ =\frac{365·364·....354}{{365}^{12}} \\ ≈0.833 \end{gathered}$$

Use the formula $P\left(\overline{E}\right)=1-P\left(E\right)$to find the probability that at least 2 people have the same birthday in a group of 12 people.

$$\begin{gathered} P\left(\overline{E}\right)=1-P\left(E\right) \\ P\left(E\right)=1-P\left(\overline{E}\right) \\ P\left(E\right)=1-0.833 \\ P\left(E\right)=0.167 \end{gathered}$$