From a set of $n$randomly chosen people, let $E_{ij}$denote the event that persons $i$and $j$have the same birthday. Assume that each person is equally likely to have any of the 365 days of the year as his or her birthday. Find

(a) $P\left(E_{3,4}\mid E_{1,2}\right)$;

(b) $P\left(E_{1,3}\mid E_{1,2}\right)$;

(c) $P\left(E_{2,3}\mid E_{1,2}\cap E_{1,3}\right)$.

What can you conclude from your answers to parts (a)-(c) about the independence of the $\left(\begin{array}{l}n\\ 2\end{array}\right)$events $E_{ij}$?

Given in the question that From a set of n randomly chosen people, let $Eij$denote the event that persons$i$and$j$ have the same birthday. Assume that each person is equally likely to have any of the $365$days of the year as his or her birthday

Keep that events $E_{1,2}$and $E_{3,4}$are separated. Understanding the information whether the first and the second have common birthdays does not influence the probabilities for the contest between the third and the fourth person. Hence

$P\left(E_{3,4}\mid E_{1,2}\right)=P\left(E_{3,4}\right)=\frac{1}{365}$

$P\left(E_{3,4}\mid E_{1,2}\right)=$$\frac{1}{365}$

From a set of n randomly chosen people, let $Eij$ denote the event that persons$i$ and $j$have the same birthday. Assume that each person is equally likely to have any of the $365$ days of the year as his or her birthday

Here we also include that events $E_{1,2}$and $E_{1,3}$are separated. This is because understanding the information whether the first or the second person match leaves chances for such a match between the first and the third person untouched. Hence

$P\left(E_{1,3}\mid E_{1,2}\right)=$$\frac{1}{365}$

From a set of n randomly chosen people, let $Eij$denote the event that persons $i$and $j$have the same birthday. Assume that each person is equally likely to have any of the $365$days of the year as his or her birthday.

Here are the circumstances a bit different. If we know that the first and the second person have the identical birthday and if we know that the first one and the third person have the same birthday, by the transitivity of that relation, there has to be that the second and the third person have the same birthday. Hence

$P\left(E_{2,3}\mid E_{1,2}\cap E_{1,3}\right)=1\ne \frac{1}{365}=P\left(E_{2,3}\right)$

$P\left(E_{2,3}\mid E_{1,2}\cap E_{1,3}\right)=$$P\left(E_{2,3}\right)$