$2.(36$ points$)$ It is said that when there are $20$ people in a room, the possibility that $2$ of them have the same

birthday $($we consider months and days only, not the years$)$ is pretty high, almost around $0\mathrm{.}5,$ at least

bigger than $0\mathrm{.}1.$

$($a$)(12$ points$)$ From the theory of probability, we know that such probability $($consider ordinary year with $365$

days$)$ is equal to $1-$ P $(365,20)/36520.$ Here P $(365,20)$ is the number of permutations of choosing $20$ from

$365$ or $365!/(345!),$ where $365!$ Is $365$ factorial or $365*364*363*362*,..*3*2*1.$

Prove this statement is correct or invalidate this by computing such a number $($using C #programming$)$

$($hint: compute $(365\mathrm{.}0/365.0)*(364\mathrm{.}0/365\mathrm{.}0)*(363\mathrm{.}0/365\mathrm{.}0)*...*(346\mathrm{.}0/365\mathrm{.}0),$ which is a for loop of

$20$ iterations, for i $=1$ to $20).$

$($b$)(12$ points$)$ Generalize this question to compute the probability of n people with $2$ of them having the same

birth date. Print out in tabular format, such probability for n $=11,12,$ up to n $=50,$ listing $5$ in a row if possible.

For verification purpose, the output should look like below:

Printing out probabilities of $2$ identical birthdays for $11$ to $50$ people

$0\mathrm{.}1411410\mathrm{.}1670250\mathrm{.}194410\mathrm{.}2231030\mathrm{.}252901$

$0\mathrm{.}2836040\mathrm{.}3150080\mathrm{.}3469110\mathrm{.}3791190\mathrm{.}411438$

$0\mathrm{.}4436880\mathrm{.}4756950\mathrm{.}5072970\mathrm{.}5383440\mathrm{.}5687$

$0\mathrm{.}5982410\mathrm{.}6268590\mathrm{.}6544610\mathrm{.}6809690\mathrm{.}706316$

$0\mathrm{.}7304550\mathrm{.}7533480\mathrm{.}7749720\mathrm{.}7953170\mathrm{.}814383$

$0\mathrm{.}8321820\mathrm{.}8487340\mathrm{.}8640680\mathrm{.}878220\mathrm{.}891232$

$0\mathrm{.}9031520\mathrm{.}914030\mathrm{.}923923$