##### Probability of only one CPU using the bus.

In a n-CPU shared bus system,if Z is the probability that any CPU requests the bus in a given cycle,the probability that only one CPU uses the bus is given by-

Options :

(1) nZ(1-Z)^{(n-1)}

(2) Z(1-Z)^{(n-1)}

(3) nZ(1-Z)^{n}

(4) (n-1)Z(1-Z)^{n}.

check with the answer.

Let us say the CPUs are C

_{1}, C_{2}, ... C_{n}.the probability that only one CPU uses the bus

= P((C

_{1}requests the bus) & (except C_{1}, other CPUs do not request the bus)) +P((C

_{2}requests the bus) & (except C_{2}, other CPUs do not request the bus)) +.....

^{ }= nZ(1-Z)^{n-1}Great explanation sir!!! Thanks

EDIT: Above explanation is correct. Thanks Arul.

Correct answer is (1).

SIMPLY THINK.....this way ..

the probability will be Z(1-Z)n-1 ...that is one get bus and others dont ..

now this one can can be any CPU=CPU1 or CPU2 or ...CPUN

=NZ(1-Z)n-1