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Math Help - Probabilities on different spots

  1. #1
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    Probabilities on different spots

    Hello,

    i've been trying to solve the following problem. If i have a system that is composed by 3 machines and all of them are needed for the super system to work, which is the probability of the system failing if each machine has a failure probability of 0.1?

    I've thought about adding all three, which would result in 0.1+0.1+0.1=0.3, but then i thought that if the probability of failure is 1, i would get 1+1+1=3, which goes against the concept of probability. Can someone help me?

    Thnks in advance
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  2. #2
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    Re: Probabilities on different spots

    Quote Originally Posted by DarkFalz View Post
    I've thought about adding all three, which would result in 0.1+0.1+0.1=0.3, but then i thought that if the probability of failure is 1, i would get 1+1+1=3, which goes against the concept of probability. Can someone help me?
    You can only add probabilities like that if the events are disjoint. If they all have the probability 1, they are most definitely not disjoint and are in fact the same event.

    If the three systems are independent, the probability of at least one of them failing is P(\text{"Failure"})=1-(1-p)^3 where p=0.1
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  3. #3
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    Re: Probabilities on different spots

    that formula means, the total probability minus the probability of all the 3 systems working well at the same time?
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  4. #4
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    Re: Probabilities on different spots

    by the way, the problem of adding all, in other words, is that i was considering the probabilty of a single system, now the impact on the three as a whole, right?
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  5. #5
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    Re: Probabilities on different spots

    Quote Originally Posted by DarkFalz View Post
    If i have a system that is composed by 3 machines and all of them are needed for the super system to work, which is the probability of the system failing if each machine has a failure probability of 0.1?
    In order for the super system to work each of the three machines must not fail.
    We assume that failure is independent. The probability the system works is (0.9)^3.
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