Probability Of Machine Working












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A complex machine is able to work if at least 3 of it’s 5 components work. If each motor independently functions for a random amount of time with density given by $f(x) = frac{x}{e^x} , x>0$, compute the density function of the length of time that the machine functions.



My approach :



Let $p$ be the probability that component $i$ works for an amount of time $t$, I can find this by integrating the pdf over $0$ to $t$.



Now probability that $P(text{at least 3 work}) = 1-P(4)^c - P(5)^c$ this I can find by assuming the above as a binomial variable.



Will this be the pdf, which is to say, is the pdf equal to $P(text{at least 3 work})$?










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    A complex machine is able to work if at least 3 of it’s 5 components work. If each motor independently functions for a random amount of time with density given by $f(x) = frac{x}{e^x} , x>0$, compute the density function of the length of time that the machine functions.



    My approach :



    Let $p$ be the probability that component $i$ works for an amount of time $t$, I can find this by integrating the pdf over $0$ to $t$.



    Now probability that $P(text{at least 3 work}) = 1-P(4)^c - P(5)^c$ this I can find by assuming the above as a binomial variable.



    Will this be the pdf, which is to say, is the pdf equal to $P(text{at least 3 work})$?










    share|cite|improve this question



























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      A complex machine is able to work if at least 3 of it’s 5 components work. If each motor independently functions for a random amount of time with density given by $f(x) = frac{x}{e^x} , x>0$, compute the density function of the length of time that the machine functions.



      My approach :



      Let $p$ be the probability that component $i$ works for an amount of time $t$, I can find this by integrating the pdf over $0$ to $t$.



      Now probability that $P(text{at least 3 work}) = 1-P(4)^c - P(5)^c$ this I can find by assuming the above as a binomial variable.



      Will this be the pdf, which is to say, is the pdf equal to $P(text{at least 3 work})$?










      share|cite|improve this question















      A complex machine is able to work if at least 3 of it’s 5 components work. If each motor independently functions for a random amount of time with density given by $f(x) = frac{x}{e^x} , x>0$, compute the density function of the length of time that the machine functions.



      My approach :



      Let $p$ be the probability that component $i$ works for an amount of time $t$, I can find this by integrating the pdf over $0$ to $t$.



      Now probability that $P(text{at least 3 work}) = 1-P(4)^c - P(5)^c$ this I can find by assuming the above as a binomial variable.



      Will this be the pdf, which is to say, is the pdf equal to $P(text{at least 3 work})$?







      probability probability-distributions random-variables binomial-distribution






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      edited Dec 3 '18 at 14:25









      Aaron Montgomery

      4,712523




      4,712523










      asked Dec 3 '18 at 14:03









      user601297

      1176




      1176






















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          I think you're close. At time $t$, the probability that a given component still works is $$p(t) = int_0^t tau e^{-tau}, dtau = left. -(tau+1) e^{-tau}right|_{tau=0}^{tau=t} = 1 - (t+1) e^{-t}.$$



          The chance that exactly $n$ out of 5 components still work at time $t$ is thus $$binom{5}{n} p(t)^n (1-p(t))^{5-n}$$
          and the chance that the machine works is just the sum of the above expression from $n = 3$ to $n = 5$, which is probably a bit messy but straightforward to compute.






          share|cite|improve this answer























          • Ok got it, thanks
            – user601297
            Dec 3 '18 at 20:07











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          I think you're close. At time $t$, the probability that a given component still works is $$p(t) = int_0^t tau e^{-tau}, dtau = left. -(tau+1) e^{-tau}right|_{tau=0}^{tau=t} = 1 - (t+1) e^{-t}.$$



          The chance that exactly $n$ out of 5 components still work at time $t$ is thus $$binom{5}{n} p(t)^n (1-p(t))^{5-n}$$
          and the chance that the machine works is just the sum of the above expression from $n = 3$ to $n = 5$, which is probably a bit messy but straightforward to compute.






          share|cite|improve this answer























          • Ok got it, thanks
            – user601297
            Dec 3 '18 at 20:07
















          0














          I think you're close. At time $t$, the probability that a given component still works is $$p(t) = int_0^t tau e^{-tau}, dtau = left. -(tau+1) e^{-tau}right|_{tau=0}^{tau=t} = 1 - (t+1) e^{-t}.$$



          The chance that exactly $n$ out of 5 components still work at time $t$ is thus $$binom{5}{n} p(t)^n (1-p(t))^{5-n}$$
          and the chance that the machine works is just the sum of the above expression from $n = 3$ to $n = 5$, which is probably a bit messy but straightforward to compute.






          share|cite|improve this answer























          • Ok got it, thanks
            – user601297
            Dec 3 '18 at 20:07














          0












          0








          0






          I think you're close. At time $t$, the probability that a given component still works is $$p(t) = int_0^t tau e^{-tau}, dtau = left. -(tau+1) e^{-tau}right|_{tau=0}^{tau=t} = 1 - (t+1) e^{-t}.$$



          The chance that exactly $n$ out of 5 components still work at time $t$ is thus $$binom{5}{n} p(t)^n (1-p(t))^{5-n}$$
          and the chance that the machine works is just the sum of the above expression from $n = 3$ to $n = 5$, which is probably a bit messy but straightforward to compute.






          share|cite|improve this answer














          I think you're close. At time $t$, the probability that a given component still works is $$p(t) = int_0^t tau e^{-tau}, dtau = left. -(tau+1) e^{-tau}right|_{tau=0}^{tau=t} = 1 - (t+1) e^{-t}.$$



          The chance that exactly $n$ out of 5 components still work at time $t$ is thus $$binom{5}{n} p(t)^n (1-p(t))^{5-n}$$
          and the chance that the machine works is just the sum of the above expression from $n = 3$ to $n = 5$, which is probably a bit messy but straightforward to compute.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 20:42

























          answered Dec 3 '18 at 15:23









          Connor Harris

          4,303723




          4,303723












          • Ok got it, thanks
            – user601297
            Dec 3 '18 at 20:07


















          • Ok got it, thanks
            – user601297
            Dec 3 '18 at 20:07
















          Ok got it, thanks
          – user601297
          Dec 3 '18 at 20:07




          Ok got it, thanks
          – user601297
          Dec 3 '18 at 20:07


















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