# Thread: System of 5 independent components

1. ## System of 5 independent components

I have done problems similar to this one with 4 components but im getting stumped with the 5th one in the middle. I thought it would be a good idea to make each side B1 and B2 to simplify the system or something like that...
The probability of failure for each components is next to the number

i basically need to find the probability that the system works..but the 5th component is confusing me
so far I have system works= (A1UA4)UA2U(A3UA5), but this is after to subtract 1 from each prob. of failure to get success.

help would be appreciated

2. Hello, mightydog78!

The probability of failure for each component is next to the number.

Find the probability that the system works.

$\text{We are given these probabiities of failure: }\;\begin{Bmatrix}P(\#1) &-& 0.01 \\ P(\#2) &=& 0.02 \\ P(\#3) &=& 0.03 \\ P(\#4) &=& 0.04 \\ P(\#5) &=& 0.05 \end{Bmatrix}$

I found four cases in which the system fails.

$\begin{array}{cccccccccc}
\text{\#1 and 4 fail:} & P(1 \wedge 4) &=& (0.01)(0.04) &=& 0.000400 \\
\text{\#3 and 5 fail:} & P(3 \wedge 5) &=& (0.03)(0.05) &=& 0.001500 \\
\text{\#1, 2 and 5 fail:} & P(1 \wedge 2 \wedge 5) &=& (0.01)(0.02)(0.05) &=& 0.000010 \\ \text{\#2, 3, and 4 fail:} & P(2 \wedge 3 \wedge 4) &=& (0.02)(0.03)(0.04) &=& 0.000024 \\ \hline &&& \text{Total:} & & 0.001934 \end{array}$

Hence: . $P(\text{system fails}) \:=\:0.001934$

Therefore: . $P(\text{system works}) \;=\; 1 - 0.001934 \;=\,0.998066$

3. hmm i was trying something similar and got a lower probability of success. how could you represent this system in terms of unions U and intersections ∩ ??