Mixing
Stuck at defining the parameter λ for an exponential distribution
$begingroup$
So I'm given this assignment in probability that states to the following:
A computer is made out of m subsystems, by which each of them has the same exponential distribution for the time of failure. The subsystems are independent from each other and the system will fail if one of the subsystems fails.
-Find the distribution function for the time of failure of the system.
-If the average time of failure for each subsystem is 800 hours and there are 5 subsystems find the probability that the time of failure is greater than 150 hours.
Now for the first one I assume that since each subsystem has an exponential distribution with the same parameter the if $Mi$ is an event that a system fails then the probability for the whole system failing P(M) is:
P(M) = P(M1) + P(M2) + P(M3) + .... + P(Mm)
or the probability density function for the time of failure of the system is:
$ p(x) = mλe^{-λx}$
and we just integrate the function.
Is that all because this seems pretty simple, or am I missing something?
And for the second one I'm having a hard time deciding. Since the average time is 800 hours is that the expected value and λ = $frac{1}{800}$ or is that just λ and λ = 800?
probability
asked Jan 27 at 22:42
David MasonDavid Mason
587
$endgroup$
add a comment |
$begingroup$
So I'm given this assignment in probability that states to the following:
A computer is made out of m subsystems, by which each of them has the same exponential distribution for the time of failure. The subsystems are independent from each other and the system will fail if one of the subsystems fails.
-Find the distribution function for the time of failure of the system.
-If the average time of failure for each subsystem is 800 hours and there are 5 subsystems find the probability that the time of failure is greater than 150 hours.
Now for the first one I assume that since each subsystem has an exponential distribution with the same parameter the if $Mi$ is an event that a system fails then the probability for the whole system failing P(M) is:
P(M) = P(M1) + P(M2) + P(M3) + .... + P(Mm)
or the probability density function for the time of failure of the system is:
$ p(x) = mλe^{-λx}$
and we just integrate the function.
Is that all because this seems pretty simple, or am I missing something?
And for the second one I'm having a hard time deciding. Since the average time is 800 hours is that the expected value and λ = $frac{1}{800}$ or is that just λ and λ = 800?
probability
asked Jan 27 at 22:42
David MasonDavid Mason
587
$endgroup$
add a comment |
$begingroup$
So I'm given this assignment in probability that states to the following:
A computer is made out of m subsystems, by which each of them has the same exponential distribution for the time of failure. The subsystems are independent from each other and the system will fail if one of the subsystems fails.
-Find the distribution function for the time of failure of the system.
-If the average time of failure for each subsystem is 800 hours and there are 5 subsystems find the probability that the time of failure is greater than 150 hours.
Now for the first one I assume that since each subsystem has an exponential distribution with the same parameter the if $Mi$ is an event that a system fails then the probability for the whole system failing P(M) is:
P(M) = P(M1) + P(M2) + P(M3) + .... + P(Mm)
or the probability density function for the time of failure of the system is:
$ p(x) = mλe^{-λx}$
and we just integrate the function.
Is that all because this seems pretty simple, or am I missing something?
And for the second one I'm having a hard time deciding. Since the average time is 800 hours is that the expected value and λ = $frac{1}{800}$ or is that just λ and λ = 800?
probability
asked Jan 27 at 22:42
David MasonDavid Mason
587
$endgroup$
So I'm given this assignment in probability that states to the following:
A computer is made out of m subsystems, by which each of them has the same exponential distribution for the time of failure. The subsystems are independent from each other and the system will fail if one of the subsystems fails.
-Find the distribution function for the time of failure of the system.
-If the average time of failure for each subsystem is 800 hours and there are 5 subsystems find the probability that the time of failure is greater than 150 hours.
Now for the first one I assume that since each subsystem has an exponential distribution with the same parameter the if $Mi$ is an event that a system fails then the probability for the whole system failing P(M) is:
P(M) = P(M1) + P(M2) + P(M3) + .... + P(Mm)
or the probability density function for the time of failure of the system is:
$ p(x) = mλe^{-λx}$
and we just integrate the function.
Is that all because this seems pretty simple, or am I missing something?
And for the second one I'm having a hard time deciding. Since the average time is 800 hours is that the expected value and λ = $frac{1}{800}$ or is that just λ and λ = 800?
probability
probability
asked Jan 27 at 22:42
David MasonDavid Mason
587
asked Jan 27 at 22:42
David MasonDavid Mason
587
asked Jan 27 at 22:42
David MasonDavid Mason
587
asked Jan 27 at 22:42
David MasonDavid Mason
587
asked Jan 27 at 22:42
David MasonDavid Mason
587
587
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $X_1, ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
{Xleq x} text{occurs if and only if} {X_1leq x} text{or} ldots text{or} {X_Mleq x} text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or ${Xgeq x}$. We have that:
$$
{Xgeq x} text{occurs if and only if} {X_1geq x} text{and} ldots text{and} {X_Mgeq x} text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X ge x)=prod_{i=1}^M P(X_i ge x) = prod_{i=1}^M e^{- lambda x} = e^{- lambda M x}
$$
so that:
$$
P(X le x) = 1- P(X ge x) = 1-e^{- lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $lambda M$.
The parameter $lambda$ is defined as the number of failures per unit time. Hence, we want $lambda = frac{1}{800}$. Therefore the probability is:
$$
P(X ge 150) = 1-P(X le 150) = 1- e^{-frac{1}{800} cdot 5 cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
answered Jan 28 at 0:17
user321627user321627
972515
$endgroup$
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $X_1, ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
{Xleq x} text{occurs if and only if} {X_1leq x} text{or} ldots text{or} {X_Mleq x} text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or ${Xgeq x}$. We have that:
$$
{Xgeq x} text{occurs if and only if} {X_1geq x} text{and} ldots text{and} {X_Mgeq x} text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X ge x)=prod_{i=1}^M P(X_i ge x) = prod_{i=1}^M e^{- lambda x} = e^{- lambda M x}
$$
so that:
$$
P(X le x) = 1- P(X ge x) = 1-e^{- lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $lambda M$.
The parameter $lambda$ is defined as the number of failures per unit time. Hence, we want $lambda = frac{1}{800}$. Therefore the probability is:
$$
P(X ge 150) = 1-P(X le 150) = 1- e^{-frac{1}{800} cdot 5 cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
answered Jan 28 at 0:17
user321627user321627
972515
$endgroup$
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
add a comment |
$begingroup$
Let $X_1, ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
{Xleq x} text{occurs if and only if} {X_1leq x} text{or} ldots text{or} {X_Mleq x} text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or ${Xgeq x}$. We have that:
$$
{Xgeq x} text{occurs if and only if} {X_1geq x} text{and} ldots text{and} {X_Mgeq x} text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X ge x)=prod_{i=1}^M P(X_i ge x) = prod_{i=1}^M e^{- lambda x} = e^{- lambda M x}
$$
so that:
$$
P(X le x) = 1- P(X ge x) = 1-e^{- lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $lambda M$.
The parameter $lambda$ is defined as the number of failures per unit time. Hence, we want $lambda = frac{1}{800}$. Therefore the probability is:
$$
P(X ge 150) = 1-P(X le 150) = 1- e^{-frac{1}{800} cdot 5 cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
answered Jan 28 at 0:17
user321627user321627
972515
$endgroup$
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
add a comment |
$begingroup$
Let $X_1, ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
{Xleq x} text{occurs if and only if} {X_1leq x} text{or} ldots text{or} {X_Mleq x} text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or ${Xgeq x}$. We have that:
$$
{Xgeq x} text{occurs if and only if} {X_1geq x} text{and} ldots text{and} {X_Mgeq x} text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X ge x)=prod_{i=1}^M P(X_i ge x) = prod_{i=1}^M e^{- lambda x} = e^{- lambda M x}
$$
so that:
$$
P(X le x) = 1- P(X ge x) = 1-e^{- lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $lambda M$.
The parameter $lambda$ is defined as the number of failures per unit time. Hence, we want $lambda = frac{1}{800}$. Therefore the probability is:
$$
P(X ge 150) = 1-P(X le 150) = 1- e^{-frac{1}{800} cdot 5 cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
answered Jan 28 at 0:17
user321627user321627
972515
$endgroup$
Let $X_1, ldots, X_M$ be the time to failure of each of the $M$ components. Then let $X$ be the time to failure of the entire system. Now if one component fails, the entire system fails. Hence, we have that the event:
$$
{Xleq x} text{occurs if and only if} {X_1leq x} text{or} ldots text{or} {X_Mleq x} text{occurs}
$$
These are rather nasty to compute so instead we can look at the complement, or ${Xgeq x}$. We have that:
$$
{Xgeq x} text{occurs if and only if} {X_1geq x} text{and} ldots text{and} {X_Mgeq x} text{occurs}
$$
Each $X_i$ is an independent exponential so:
$$
P(X ge x)=prod_{i=1}^M P(X_i ge x) = prod_{i=1}^M e^{- lambda x} = e^{- lambda M x}
$$
so that:
$$
P(X le x) = 1- P(X ge x) = 1-e^{- lambda M x}
$$
which is the CDF of an exponential distribution with rate parameter $lambda M$.
The parameter $lambda$ is defined as the number of failures per unit time. Hence, we want $lambda = frac{1}{800}$. Therefore the probability is:
$$
P(X ge 150) = 1-P(X le 150) = 1- e^{-frac{1}{800} cdot 5 cdot 150}
$$
One helpful way to always ensure you have it correct is to make sure that the time units in the exponent cancel out.
answered Jan 28 at 0:17
user321627user321627
972515
edited Jan 28 at 0:47
answered Jan 28 at 0:17
user321627user321627
972515
answered Jan 28 at 0:17
user321627user321627
972515
answered Jan 28 at 0:17
user321627user321627
972515
972515
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
add a comment |
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
Why is the first one nasty to compute? Since {X≤x} occurs if and only if {X1≤x} or … or {XM≤x} occurs the P{ X < x} = $sum_{i=1}^{m} 1 - e^{-λx} $ which is equaled to $m(1 - e^{-λx})$ which doesn't add up with your CDF. Can you tell me what I'm doing wrong?
$endgroup$
– David Mason
Jan 28 at 18:57
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
$endgroup$
– user321627
Jan 28 at 21:42
$begingroup$
When dealing with the "or's", we have to consider many many cases. ${Xleq x}$ refers to the event the entire system fails before time $x$. One possible scenario is $X_1$ failing before time $x$, and $X_2, ldots, X_M$ not failing. Another scenario is $X_1$ not failing before time $x$, and $X_3, ldots, X_M$ not failing up to time $x$, BUT $X_2$ failing before time $x$. Both these scenarios will cause the system to fail before time $x$, and so we have to add ALL the possible cases in which this might arise. Taking into account all the scenarios requires a lot of summations.
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– user321627
Jan 28 at 21:42
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
The sum you have refers only to one such possible scenario for the entire system to fail. Specifically, what you computed is the case when ALL $X_1, ldots, X_M$ have failed before time $x$. But the problem states that for the entire system to fail, we don't need all subsystems to fail, just at least one needs to fail. In short, there are many many possible combinations/situations for the entire system to fail, and if we want to compute the probability we have to sum over all the possible cases, which will require combinatorics over sums.
$endgroup$
– user321627
Jan 28 at 21:47
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
$begingroup$
I might be able to take a stab at where your confusion might lie. In general, $P(A text{or} B text{or} C) = P(A) + P(B) + P(C)$ only when the events $A,B,C$ are mutually exclusive. However, here there is no mention of mutual exclusion here, only independence. In fact, the events are not mutually exclusive, because it were the case, then it would imply that one subsystem failing before time $x$ means that another subsystem cannot fail before time $x$ as well, which is not true.
$endgroup$
– user321627
Jan 28 at 21:56
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