Probability Of Machine Working
$begingroup$
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
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
$endgroup$
add a comment |
$begingroup$
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
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
$endgroup$
add a comment |
$begingroup$
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
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
$endgroup$
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
probability probability-distributions random-variables binomial-distribution
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
edited Dec 3 '18 at 14:25
Aaron Montgomery
4,732523
4,732523
asked Dec 3 '18 at 14:03
user601297user601297
39219
asked Dec 3 '18 at 14:03
user601297user601297
39219
asked Dec 3 '18 at 14:03
user601297user601297
39219
39219
add a comment |
add a comment |
1 Answer
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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.
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
$endgroup$
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
$endgroup$
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
$endgroup$
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
add a comment |
$begingroup$
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.
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
$endgroup$
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.
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
edited Dec 3 '18 at 20:42
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
answered Dec 3 '18 at 15:23
Connor HarrisConnor Harris
4,355724
4,355724
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
add a comment |
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
$begingroup$
Ok got it, thanks
$endgroup$
– user601297
Dec 3 '18 at 20:07
add a comment |
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