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Math Help - System of 5 independent components

  1. #1
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    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
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  2. #2
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    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}<br />
\text{\#1 and 4 fail:} & P(1 \wedge 4) &=& (0.01)(0.04) &=& 0.000400 \\<br />
\text{\#3 and 5 fail:} & P(3 \wedge 5) &=& (0.03)(0.05) &=& 0.001500 \\<br />
\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
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  3. #3
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    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 ∩ ??
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