Thread: Probability System of four independent components

Here is a probability system


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

Originally Posted by mightydog78

Here is a probability system


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) = ....

Hello, mightydog78!

Here is a probability system


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.

. .

(a) The system works if:

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

. . . .


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

. . . .


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

. . . .


Therefore: .


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


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

. .


Therefore: .


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





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

. .
. .

Hence, the numerator is: .

And we found the denominator in part (b): .


Therefore: .

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.