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Math Help - Probability System of four independent components

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by mightydog78 View Post
    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) = ....
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  3. #3
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    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.

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

    (a) The system works if:

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

    . . . . 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.

    . . . . 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.

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


    Therefore: . 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."

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


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


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


    (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):

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

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

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


    Therefore: . P(\text{system works }|\text{ at most 3 work}) \;=\;\dfrac{0.2772}{0.6976} \;=\;\dfrac{693}{1744}
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    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.
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