The number of arrangements in which all the letters of the given words can be arranged = $\frac{13 !}{4 ! 3 !2! 2 !}$
(a) When all four S's come together, then the no. of arrangement = $\frac{10 !}{3 ! 2 ! 2 !}$
$\therefore $ Required probability = $\frac{\frac{10!}{3 ! \;2!\;2!}}{\frac{13!}{4!\; 3!\; 2!\; 2!}} = \frac{10! 4!}{13!}$
$\; \; \; \; \; \; \; \; = \frac{2}{143}$
(b) The no. of arrangements in which all A's come together = $\frac{11!}{ 4! 2! 2!}$
$\therefore $ Probability of all A's coming together = $\frac{\frac{11!}{4! \; 2! \; 2!}}{\frac{13!}{4! \; 3! \; 2!\; 2!}} = \frac{11! \times 3!}{13!}$
$\; \; \; \; \; \; \; \; \; \; = \frac{1}{26}$
(c) The no. of arrangement in which two I's and two N's come together = $\frac{10!}{4! \; 3!} \times \frac{4!}{2! \times 2!} = \frac{10!}{3! \;2! \; 2!}$
$\therefore $ Required probability = $\frac{\frac{10!}{3! \; 2! \; 2!}}{ \frac{13!}{4! \; 3! \ 2! \;2! }} = \frac{10! \times 4 !}{13!}$
$ \; \; \; \; \; \; \; \; \; \; \; = \frac{2}{124}$