chefren wrote:
Two equal power supplies "in line": Twice as much the risk of a
brakedown of the system and two times as much failures of power supplies.
Lets see.
Let X be the (boolean) random variable designating ''system X breaks
down in the first N years''. Equally, let Y be the random variable
designating ''system Y breaks down in the first N years''.
Then P(X = 1) is the probability of X breaking down and similarly, P(Y =
1) is the probability Y breaks down.
Now (X = 1) and (Y = 1) are clearly independent. If one breaks down, it
does not influence wether or not the other one does. But since the
events are independent, they cannot be mutually exclusive. This makes
sense logically, since both X and Y can break down in N years so
intecsection(X = 1, Y = 1) is not the empty set which implies X and Y
not mutually exclusive.
The addition rule for independent events gives us:
P(union(X = 1, Y = 1)) = P(X = 1) + P(Y = 1) - P(X = 1) * P(Y = 1)
So you forget the last term by saying ''twice as much''. You have to
deduct the probability that both events occur (or it would have been
''counted'' twice).
Two equal power supplies in parallel: Half the risk of a brakedown of
the system but still two times as much failures of power supplies and
twice the support effort for the power supplies.
Now in this case, we still have independence, but now both has to fail.
In other words
P(intersection(X = 1, Y = 1)) = P(X = 1) * P(Y = 1)
This is theory. In practice a failing power supply will be changed as
soon as it shows an error. Especially in the serial case. This means
that in practice, one has to do a more heavyweight probability analysis.
One needs the probabilities after one month, after 2, 3, 4, etc to do
the discrete case. I can assure you the probabilities are not as easy as
you are taking them to be.
