if n units are independent and the probability of each being up is 2/3 then the probability of all n being up is (2/3)^n
HOWEVER I can not see your flow chart and therefore can not evaluate parallel paths
if there are two ways to get from A to B
and the probability of failure of each branch is 1/3
then the probability of both failing is (1/3)(1/3) = 1/9 and the probability of getting from A to B one way or another is 1 - 1/9 = 8/9
Suppose that each unit of a system is up with probability 2/3 and down with probability 1/3. Different units are independent. For each one of the systems shown below, calculate the probability that the whole system is up (that is, that there exists a path from the left end to the right end, consisting entirely of units that are up).
What is the probability that the following system is up?
What is the probability that the following system is up?
6 answers
With f the probability of fail
a) First branch: f1=(1/3)*(1/3)
Second branch: f2=1/3
Thus:
The probability of success of the system is:(1-(1/3)*(1/3))*2/3
=2/3*(1-1/9)=(2*8)/(3*9)
b)
First branch:
f1= 1-(2/3)*(2/3)=5/9
second branch
f2=1/3
Thus:
The probability of success of the system is: 1-(5/9)*(1/3)
a) First branch: f1=(1/3)*(1/3)
Second branch: f2=1/3
Thus:
The probability of success of the system is:(1-(1/3)*(1/3))*2/3
=2/3*(1-1/9)=(2*8)/(3*9)
b)
First branch:
f1= 1-(2/3)*(2/3)=5/9
second branch
f2=1/3
Thus:
The probability of success of the system is: 1-(5/9)*(1/3)
a)diagrams
-p1 -(p2)- end
-(p3)- end
b)-(p1)-(p2)- end
-(p3)----- end
-p1 -(p2)- end
-(p3)- end
b)-(p1)-(p2)- end
-(p3)----- end
probability of a system being up in series = p1*p2*...pn
probability of a system being up in parallel=1 -((1-p1)*(1-p2)*..(1-pn)
where pn is the probability of a unit
so in this case p(Up)=p[1 -((1-p)*(1-p)]
p(Up)=2/3*(1-((1/3)(1/3))=18/27
probability of a system being up in parallel=1 -((1-p1)*(1-p2)*..(1-pn)
where pn is the probability of a unit
so in this case p(Up)=p[1 -((1-p)*(1-p)]
p(Up)=2/3*(1-((1/3)(1/3))=18/27
then part b we do the same the top part is in series
Top part=[p*p]=2/3*2/3=4/9
the bottom part=2/3
and we the apply this formula as they are now in parallel
probability of a system being up in parallel=1 -((1-p1)*(1-p2)*..(1-pn)
in this case it will be 1-[5/9*2/3]=17/27
Top part=[p*p]=2/3*2/3=4/9
the bottom part=2/3
and we the apply this formula as they are now in parallel
probability of a system being up in parallel=1 -((1-p1)*(1-p2)*..(1-pn)
in this case it will be 1-[5/9*2/3]=17/27
In the first diagram, the parallel connection of the two units (on the right) is down when both units fail, which happens with probability (1/3)⋅(1/3)=1/9. Therefore the parallel connection is up with probability 1−1/9=8/9. The overall system is up if the first unit is up (probability 2/3) and the parallel connection is also up (probability 8/9), which happens with probability (8/9)⋅(2/3)=16/27.
In the second diagram, the top path is up when both of its units are up – this happens with probability (2/3)⋅(2/3)=4/9. Thus it fails with probability 1−4/9=5/9. The overall system fails when the top path fails (probability 5/9) and the bottom path also fails (probability 1/3). Thus the probability of failure is (5/9)⋅(1/3)=5/27. It follows that the probability that the system is up (does not fail) is 1−5/27=22/27.
In the second diagram, the top path is up when both of its units are up – this happens with probability (2/3)⋅(2/3)=4/9. Thus it fails with probability 1−4/9=5/9. The overall system fails when the top path fails (probability 5/9) and the bottom path also fails (probability 1/3). Thus the probability of failure is (5/9)⋅(1/3)=5/27. It follows that the probability that the system is up (does not fail) is 1−5/27=22/27.
41 answers