Solution

a. Let $E_{xy}$ be the event that first and second person has birthday on date $x$ and $y$, respectively.

The simple events would be of the form above and $365^2$ (Simplification: assuming that all years have 365 days.)

b. Each simple event is equally likely. Thus, probability to have specific unordered pair $(x,y)$ is given by

​

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C.The number of simple events where both person have same birthday is given by 

$C_1^{365}= 365.$

$$A= \{ E_{ii}: i \in \{1, \dots, 365\} \}$$

d. Since, in this simplified problem, all the simple events are equally likely,

$$\begin{align*} P(A)&= \dfrac{\textrm{No. of simple events in $A$}}{\textrm{Total number of simple events}}\\ &= \dfrac{365}{365^2}\\ &= \dfrac{1}{365} \end{align*}$$

e. By Addition rule,

 

=365364​​

P(Ac)=. 1−P(A)​

Answer

$$$$
\begin{equation*} \textrm{b. } P(\textrm{ birthday on $x$ and $y$ })= \left\{ \begin{array}{lr} 1/(365)^2 & \textrm{ if } x=y \\ 2/(365)^2 &\textrm{ if } x \ne y \end{array} \right. \textrm{d. 1/365 ~~e. 364/365}\\ \end{equation*}