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Question on calculating probability of specific items from multiple number of items
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There is a lighting that has failure rate of 0.05/hour.
Total of 8 lightings are installed on a car.
Two on the front-right, two on the front-left, two on the rear-right and two on the rear-left. All have the same failure rate.
I'd think the probability of any 2 lightings to fail at the same time (one specific hour) is $0.05^2 = 0.0025$ per hour.
Whats confusing me is what the probability would be for the two (2) specific lightings to fail at the same time (one specific hour).
For example, what is the probability of the two front-right lightings to fail at the same time (one specific hour)?
I don't think its as simple as $0.05^2$.
probability
asked Jul 24 at 0:26
Han Kim
32
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up vote
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favorite
There is a lighting that has failure rate of 0.05/hour.
Total of 8 lightings are installed on a car.
Two on the front-right, two on the front-left, two on the rear-right and two on the rear-left. All have the same failure rate.
I'd think the probability of any 2 lightings to fail at the same time (one specific hour) is $0.05^2 = 0.0025$ per hour.
Whats confusing me is what the probability would be for the two (2) specific lightings to fail at the same time (one specific hour).
For example, what is the probability of the two front-right lightings to fail at the same time (one specific hour)?
I don't think its as simple as $0.05^2$.
probability
asked Jul 24 at 0:26
Han Kim
32
The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There is a lighting that has failure rate of 0.05/hour.
Total of 8 lightings are installed on a car.
Two on the front-right, two on the front-left, two on the rear-right and two on the rear-left. All have the same failure rate.
I'd think the probability of any 2 lightings to fail at the same time (one specific hour) is $0.05^2 = 0.0025$ per hour.
Whats confusing me is what the probability would be for the two (2) specific lightings to fail at the same time (one specific hour).
For example, what is the probability of the two front-right lightings to fail at the same time (one specific hour)?
I don't think its as simple as $0.05^2$.
probability
asked Jul 24 at 0:26
Han Kim
32
There is a lighting that has failure rate of 0.05/hour.
Total of 8 lightings are installed on a car.
Two on the front-right, two on the front-left, two on the rear-right and two on the rear-left. All have the same failure rate.
I'd think the probability of any 2 lightings to fail at the same time (one specific hour) is $0.05^2 = 0.0025$ per hour.
Whats confusing me is what the probability would be for the two (2) specific lightings to fail at the same time (one specific hour).
For example, what is the probability of the two front-right lightings to fail at the same time (one specific hour)?
I don't think its as simple as $0.05^2$.
probability
asked Jul 24 at 0:26
Han Kim
32
edited Jul 24 at 2:39
asked Jul 24 at 0:26
Han Kim
32
asked Jul 24 at 0:26
Han Kim
32
asked Jul 24 at 0:26
Han Kim
32
32
The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24
add a comment |Â
The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24
The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
It needs to be clear whether the $0.05$/hour is intended to be the chance that one specific light fails or that one of the eight lights fails. Even if it is one of the eight that means you have a failure about every $20$ hours, which is terrible.
If the $0.05/$hour is per light the chance that two specific lights fail in a particular hour is then $0.05^2$.
If the $0.05$/hour is for the whole set you need to translate that to the chance that an individual light fails in one hour. You have $0.95$ chance that all eight survive the hour, so $0.95^1/8approx 0.9936$ chance that any given one survives and $0.0064$/hour failure rate of one light. The chance that two specific lights fail in one hour is the square of that.
answered Jul 24 at 0:51
Ross Millikan
275k21186351
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It needs to be clear whether the $0.05$/hour is intended to be the chance that one specific light fails or that one of the eight lights fails. Even if it is one of the eight that means you have a failure about every $20$ hours, which is terrible.
If the $0.05/$hour is per light the chance that two specific lights fail in a particular hour is then $0.05^2$.
If the $0.05$/hour is for the whole set you need to translate that to the chance that an individual light fails in one hour. You have $0.95$ chance that all eight survive the hour, so $0.95^1/8approx 0.9936$ chance that any given one survives and $0.0064$/hour failure rate of one light. The chance that two specific lights fail in one hour is the square of that.
answered Jul 24 at 0:51
Ross Millikan
275k21186351
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
add a comment |Â
up vote
0
down vote
accepted
It needs to be clear whether the $0.05$/hour is intended to be the chance that one specific light fails or that one of the eight lights fails. Even if it is one of the eight that means you have a failure about every $20$ hours, which is terrible.
If the $0.05/$hour is per light the chance that two specific lights fail in a particular hour is then $0.05^2$.
If the $0.05$/hour is for the whole set you need to translate that to the chance that an individual light fails in one hour. You have $0.95$ chance that all eight survive the hour, so $0.95^1/8approx 0.9936$ chance that any given one survives and $0.0064$/hour failure rate of one light. The chance that two specific lights fail in one hour is the square of that.
answered Jul 24 at 0:51
Ross Millikan
275k21186351
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It needs to be clear whether the $0.05$/hour is intended to be the chance that one specific light fails or that one of the eight lights fails. Even if it is one of the eight that means you have a failure about every $20$ hours, which is terrible.
If the $0.05/$hour is per light the chance that two specific lights fail in a particular hour is then $0.05^2$.
If the $0.05$/hour is for the whole set you need to translate that to the chance that an individual light fails in one hour. You have $0.95$ chance that all eight survive the hour, so $0.95^1/8approx 0.9936$ chance that any given one survives and $0.0064$/hour failure rate of one light. The chance that two specific lights fail in one hour is the square of that.
answered Jul 24 at 0:51
Ross Millikan
275k21186351
It needs to be clear whether the $0.05$/hour is intended to be the chance that one specific light fails or that one of the eight lights fails. Even if it is one of the eight that means you have a failure about every $20$ hours, which is terrible.
If the $0.05/$hour is per light the chance that two specific lights fail in a particular hour is then $0.05^2$.
If the $0.05$/hour is for the whole set you need to translate that to the chance that an individual light fails in one hour. You have $0.95$ chance that all eight survive the hour, so $0.95^1/8approx 0.9936$ chance that any given one survives and $0.0064$/hour failure rate of one light. The chance that two specific lights fail in one hour is the square of that.
answered Jul 24 at 0:51
Ross Millikan
275k21186351
answered Jul 24 at 0:51
Ross Millikan
275k21186351
answered Jul 24 at 0:51
Ross Millikan
275k21186351
answered Jul 24 at 0:51
Ross Millikan
275k21186351
275k21186351
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
add a comment |Â
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Thanks Ross. 0.05/hr is for one specific light to fail. So if I want to know the probablitiy of two Front-Right lights to fail at the same time, its 0.05^2 = 0.0025/hr
– Han Kim
Jul 24 at 1:28
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
Yes, it is, if by same time you mean within one specific hour. They won't fail at exactly the same time.
– Ross Millikan
Jul 24 at 2:10
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
thats right, at one specific hour. Thank you
– Han Kim
Jul 24 at 2:20
add a comment |Â
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The probability that any two lights fail at the exact same time is zero, for the same reason that the probability that two independent continuous random variables are equal is zero.
– Math1000
Jul 24 at 0:47
Hi math1000, I don't fully understand how the probablity can be zero? Could you please elaborate? Thanks
– Han Kim
Jul 24 at 1:34
The wording "failure rate of 0.05/hour" suggests that the time to failure of a light is exponentially distributed with rate $1/20$. Assuming the lights are independent of each other, the probability that they fail at the exact same time is zero, since the probability that two independent continuous random variables are equal is zero. If you do not understand why that is the case, I suggest you review your notes/textbook.
– Math1000
Jul 24 at 1:37
Thanks Math1000. I shouldve mentioned that the definition of "same time" is at one specific hour. In that case, the probablitiy wouldn't be 0, correct?
– Han Kim
Jul 24 at 2:22
Yes, that completely changes the problem. I recommend you edit your question to make that clear.
– Math1000
Jul 24 at 2:24