Why this prove the independence of these events? I'm studying A First Course in Probability from  Sheldon Ross, and I'm trying to solve this exercise from page 171:
 
My solution:
a. $P(E_{3,4}|E_{1,2})= \frac{P(E_{3,4}E_{1,2})}{P(E_{1,2})}=\frac{1/365^2}{1/365}=\frac{1}{365}$
b. $P(E_{1,3}|E_{1,2})= \frac{P(E_{3,4}E_{1,2})}{P(E_{1,2})}=\frac{1/365^2}{1/365}=\frac{1}{365}$
c. $P(E_{2,3}|E_{1,2}\cap E_{1,3}) = \frac{P(E_{2,3}E_{1,2}E_{1,3})}{P(E_{1,2}E_{1,3})}=\frac{1/365^3}{1/365^2}=\frac{1}{365}$
My questions:


*

*Are my calculations right?

*Why this prove the independence of these events? can I use a similar argument to show the independence of events in general?


a. $\checkmark$ Indeed. When given that persons 1 and 2 share a birthdate, there is no influence on whether persons 3 and 4 do too. $$\mathsf P(E_{3,4}\mid E_{1,2})=\dfrac 1{365}$$
b. $\checkmark$ Indeed.  When given that persons 1 and 2 share a birthdate, there is no influence on whether person 3 shares a birthdate with person 1. $$\mathsf P(E_{1,3}\mid E_{1,2})=\dfrac 1{365}$$
c. $\times$. Nope.  When given that persons 1 and 2 share a birthdate, and that 1 and 3 do too, that makes it certain that 2 and 3 share a birth date.  $$\mathsf P(E_{2,3}\mid E_{1,2}\cap E_{1,3})=1$$
