Let n balls each bearing any number between 1 and 365 are placed in a row and a ball is selected at random. Let P be the probability that the number on the selected ball is same as the number on any one ball placed on left of it and all balls on left of it bear distinct numbers. If for P to be maximum the
position of the selected ball counting from left to right is k, then find the sum of digits of k

Let kth ball be selected, then each of k–1 balls on left of it must have a distinct number and kth ball must have a number which is on one of the k–1 balls on left. Probability that k–1 balls each bearing a distinct number between $1\mathrm{\&}365=\frac{{ }^{365}{\mathrm{C}}_{\mathrm{k}-1}\times (\mathrm{k}-1)!}{{365}^{\mathrm{k}-1}}$
Probability that the kth ball has the number on any of the numbers on k–1balls $=\frac{(\mathrm{k}-1)}{365}$Hence probability of k^th being selected
${\mathrm{P}}_{\mathrm{k}}=\frac{{ }^{365}{\mathrm{C}}_{\mathrm{k}-1}\times (\mathrm{k}-1)!}{{365}^{\mathrm{k}-1}}\times \frac{\mathrm{k}-1}{365}$
For maximum probability of ${\mathrm{P}}_{\mathrm{k}-1}\le {\mathrm{P}}_{\mathrm{k}}\ge {\mathrm{P}}_{\mathrm{k}+1}$
$\begin{array}{l}\Rightarrow \frac{(\mathrm{k}-2){\times }^{365}{\mathrm{C}}_{\mathrm{k}-2}\times (\mathrm{k}-2)!}{{365}^{\mathrm{k}-1}}\\ \le \frac{(\mathrm{k}-1){\times }^{365}{\mathrm{C}}_{\mathrm{k}-1}\times (\mathrm{k}-1)!}{{365}^{\mathrm{k}}}\ge \frac{\mathrm{k}{\times }^{365}{\mathrm{C}}_k\times \mathrm{k}!}{{365}^{\mathrm{k}+1}}\\ \Rightarrow \frac{(\mathrm{k}-2)}{(367-\mathrm{k})}\le \frac{(\mathrm{k}-1)}{365}\mathrm{\&}\frac{(\mathrm{k}-1)}{(366-\mathrm{k})}\ge \frac{\mathrm{k}}{365}\\ \Rightarrow {\mathrm{k}}^2-3\mathrm{k}-363\le 0\mathrm{\&}{\mathrm{k}}^2-\mathrm{k}-365\ge 0\\ \Rightarrow \frac{1+\sqrt{1461}}{2}\le \mathrm{k}\le \frac{3+\sqrt{1461}}{2}\text{ hence }\mathrm{k}=20\end{array}$