$6.$ If we know the number of cache hits and misses, we can calculate the average memory access time $($AMAT$)$ of the program. If we know the information of hits and misses, the AMAT of the program can be calculated as follows.

AMAT $=($Phit TM$)+($PMiss $$ TD$)$

Where TM represents the memory access cost, To represents the disk access cost, TD represents the probability of finding the data item in the cache, and PMiss represents the probability of not finding the data in the cache. PMiss and PHit have values $$between $0\mathrm{.}0$ and $1\mathrm{.}0,$ satisfying Pviss $+$ Phit $=1\mathrm{.}0.$ For example, let's say a computer has a small memory of $4$ KB and the size of the page is $256$ bytes. The virtual address consists of two parts: a $4-$bit VPN $($most significant bit$)$ and an $8-$bit offset $($least significant bit$).$ And one process can access $24$ or a total of $16$ virtual pages. The process points to memory at virtual addresses $0$x$000,0$x$100,0$x$200,0$x$300,0$x$400,0$x$500,0$x$600,0$x$700,0$x$800,$ and $0$x$900.$ These virtual addresses represent the first bytes of the first ten pages of the address space $($the first hexadecimal digit of each virtual address represents the page number$).(1)$ Assuming that all pages except virtual page $3$ are already in memory, find Pri and PMiss. $(2)$ In this case, assuming that the cost of accessing memory $($Tm$)$ is approximately $100$ nsec and the cost of accessing disk $($To$)$ is approximately $10$sec$,$ find AMAT. $(3)$ What happens to AMAT if the hit rate is $99\mathrm{.}9\%?[6$ points$]$