Suppose that there are n CPUs on a common bus. The probability that any CPU tries to use the bus in a given cycle is p. What is the chance that: a. The bus is idle (0 requests).
b. Exactly one request is made.
c. More than one request is made.

a) ( 1 - P )^n

b) nP ( 1 - p )^n-1

c) 1 - p ( 1 - p )^n-1  -  ( 1 - p )^n-1

Explanation:

A) Chance that the bus is idle

For the bus to be idle, the CPU has to be idle as well and the probability of an idle CPU is = 1 - P .therefore The chance that the bus is idle all N numbers of CPU's are idle/silent  =  ( 1 - P )^n

B) Chance that exactly one request is made

probability in the CPU when a single request is made can be expressed as :

P ( 1 - p ) ^n-1

probability that a single request is made to the Bus due to the single request made to N number of CPU's = nP ( 1 - p )^n-1

C) chance the more than one request is made

= 1 - p ( 1 - p )^n-1  -  ( 1 - p )^n-1

the answer is b) 2nd

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answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation: