# Probability System of four independent components

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• Sep 17th 2010, 04:58 PM
mightydog78
Probability System of four independent components
Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg

The probability of failure of #1=0.1, #2=0.2,#3=0.3,#4=0.4
Now im trying to find the probability that the system works..
I so far have 1∩(2U4)∩3 which seems right but I dont know what numbers to plug in and whether to add or multiply

Also, I need to find the prob. that at most 3 events of these components work.
Then, using conditional prob. I can't seem to find the prob. that the system works given that at most 3 of the components work..

Im so lost
• Sep 17th 2010, 07:34 PM
mr fantastic
Quote:

Originally Posted by mightydog78
Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg

The probability of failure of #1=0.1, #2=0.2,#3=0.3,#4=0.4
Now im trying to find the probability that the system works..
I so far have 1∩(2U4)∩3 which seems right but I dont know what numbers to plug in and whether to add or multiply

Also, I need to find the prob. that at most 3 events of these components work.
Then, using conditional prob. I can't seem to find the prob. that the system works given that at most 3 of the components work..

Im so lost

Is the working of each component assumed to be independent of the other components?

Pr(1 works) = 0.9.

Pr(2 or 4 works) = Pr(2 works) + Pr(4 works) - Pr( 2 and 4 both work) = ....

Pr(3 works) = ....
• Sep 17th 2010, 09:20 PM
Soroban
Hello, mightydog78!

Quote:

Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg

The probability of failure of #1 = 0.1, #2 = 0.2,#3 = 0.3,#4 = 0.4

(a) Now im trying to find the probability that the system works.

(b) Also, I need to find the prob. that at most 3 of these components work.

(c)Then, using conditional prob. I can't seem to find the prob. that
. . the system works, given that at most 3 of the components work.

. . $\displaystyle \begin{array}{|c||c|c|} \hline \text{Component} & \text{P(work)} & \text{P(fail)} \\ \hline \#1 & 0.9 & 0.1 \\ \#2 & 0.8 & 0.2 \\ \#3 & 0.7 & 0.3 \\ \#4 & 0.6 & 0.4 \\ \hline \end{array}$

(a) The system works if:

. . [1] #1 works, #2 works, #3 works, and #4 works.

. . . . $\displaystyle P(1,2,3,4) \:=\:(0.9)(0.8)(0.7)(0.6) \:=\:0.3024$

. . [2] #1 works, #2 works, #3 works, and #4 fails.

. . . . $\displaystyle P(1,2,3,\overline4) \:=\:(0.9)(0.8)(0.7)(0.4) \:=\:0.2016$

. . [3] #1 works, #2 fails, #3 works, and #4 works.

. . . . $\displaystyle P(1,\overline2,3,4) \:=\:(0.9)(0,2)(0.7)(0.6) \:=\:0.0756$

Therefore: .$\displaystyle P(\text{system works}) \:=\:0.3024 + 0.2016 + 0.0756 \:=\:0.5796$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(b) The opposite of "at most 3 components work" is "4 components work."

. . $\displaystyle P(\text{4 work}) \:=\:(0.9)(0.8)(0.7)(0.6) \:=\:0.3024$

Therefore: .$\displaystyle P(\text{at most 3 work}) \;=\;1 - 0.3024 \;=\;0.6976$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle (c)\;P(\text{system works }|\text{ at most 3 work}) \;=\;\dfrac{P(\text{system works }\wedge\text{ at most 3 work})}{P(\text{at most 3 work})}$

The numerator is [2] and [3] in part (a):

. . $\displaystyle [2]\;P(1,2,3,\overline4) \:=\:(0.9)(0.8)(0.7)(0.4) \:=\:0.2016$
. . $\displaystyle [3]\;P(1,\overline2,3,4) \:=\:(0.9)(0,2)(0.7)(0.6) \:=\:0.0756$

Hence, the numerator is: .$\displaystyle 0.2016 + 0.0756 \:=\:0.2772$

And we found the denominator in part (b): .$\displaystyle 0.6976$

Therefore: .$\displaystyle P(\text{system works }|\text{ at most 3 work}) \;=\;\dfrac{0.2772}{0.6976} \;=\;\dfrac{693}{1744}$
• Sep 18th 2010, 09:03 AM
mightydog78
Quote:

Originally Posted by mr fantastic
Is the working of each component assumed to be independent of the other components?

Pr(1 works) = 0.9.

Pr(2 or 4 works) = Pr(2 works) + Pr(4 works) - Pr( 2 and 4 both work) = ....

Pr(3 works) = ....

they are independent of each other but a 1 and 3 need to work for the whole system to work, and 2 or 4 must work for the system to work as well.