Find the total probability of at least one take-off distaster happening over one year
Every minute, a random object might appear on the runway of a certain airport with the probability of $p = frac {1}{10^6}$. This object will cause a take-off disaster with the probability of $r= frac{1}{10^3}$. What is the probability that at least one take-off disaster will happen over the coure of a year, given that the state of the runway is controlled every hour?
First of all, I want to find out what the hourly probabiltiy of the crash is. Every object appearing on the runway may or may not cause the crash. The probability that the plane crashes is therefore:
$$p(text{crash})= p(text{object}) cdot p(text{crash | object}) + p(text{no object}) cdot p(text{crash | no object})$$
$$p(text{crash}) = frac{1}{10^6} cdot frac{1}{10^3} + 0 = frac{1}{10^9}$$
This is the probabiliy that the plane will crash every minute.
The total probabiliy that the plane will crash over an hour is:
$$p(text{hour crash}) = 1 - left( frac{10^9 - 1}{10^9} right)^{60} = 6 cdot 10^{-8} $$
The year has $365 cdot 24 = 8760$ hours.
And so, the probabiliy that the plane does crash is:
$$p(text{year crash})= 1 - (1 -6 cdot 10^{-8})^{8760}= 0.00053$$
Do you think my solution is correct?
probability discrete-mathematics
asked Nov 28 '18 at 11:06
Aemilius
1,686314
add a comment |
Every minute, a random object might appear on the runway of a certain airport with the probability of $p = frac {1}{10^6}$. This object will cause a take-off disaster with the probability of $r= frac{1}{10^3}$. What is the probability that at least one take-off disaster will happen over the coure of a year, given that the state of the runway is controlled every hour?
First of all, I want to find out what the hourly probabiltiy of the crash is. Every object appearing on the runway may or may not cause the crash. The probability that the plane crashes is therefore:
$$p(text{crash})= p(text{object}) cdot p(text{crash | object}) + p(text{no object}) cdot p(text{crash | no object})$$
$$p(text{crash}) = frac{1}{10^6} cdot frac{1}{10^3} + 0 = frac{1}{10^9}$$
This is the probabiliy that the plane will crash every minute.
The total probabiliy that the plane will crash over an hour is:
$$p(text{hour crash}) = 1 - left( frac{10^9 - 1}{10^9} right)^{60} = 6 cdot 10^{-8} $$
The year has $365 cdot 24 = 8760$ hours.
And so, the probabiliy that the plane does crash is:
$$p(text{year crash})= 1 - (1 -6 cdot 10^{-8})^{8760}= 0.00053$$
Do you think my solution is correct?
probability discrete-mathematics
asked Nov 28 '18 at 11:06
Aemilius
1,686314
add a comment |
Every minute, a random object might appear on the runway of a certain airport with the probability of $p = frac {1}{10^6}$. This object will cause a take-off disaster with the probability of $r= frac{1}{10^3}$. What is the probability that at least one take-off disaster will happen over the coure of a year, given that the state of the runway is controlled every hour?
First of all, I want to find out what the hourly probabiltiy of the crash is. Every object appearing on the runway may or may not cause the crash. The probability that the plane crashes is therefore:
$$p(text{crash})= p(text{object}) cdot p(text{crash | object}) + p(text{no object}) cdot p(text{crash | no object})$$
$$p(text{crash}) = frac{1}{10^6} cdot frac{1}{10^3} + 0 = frac{1}{10^9}$$
This is the probabiliy that the plane will crash every minute.
The total probabiliy that the plane will crash over an hour is:
$$p(text{hour crash}) = 1 - left( frac{10^9 - 1}{10^9} right)^{60} = 6 cdot 10^{-8} $$
The year has $365 cdot 24 = 8760$ hours.
And so, the probabiliy that the plane does crash is:
$$p(text{year crash})= 1 - (1 -6 cdot 10^{-8})^{8760}= 0.00053$$
Do you think my solution is correct?
probability discrete-mathematics
asked Nov 28 '18 at 11:06
Aemilius
1,686314
Every minute, a random object might appear on the runway of a certain airport with the probability of $p = frac {1}{10^6}$. This object will cause a take-off disaster with the probability of $r= frac{1}{10^3}$. What is the probability that at least one take-off disaster will happen over the coure of a year, given that the state of the runway is controlled every hour?
First of all, I want to find out what the hourly probabiltiy of the crash is. Every object appearing on the runway may or may not cause the crash. The probability that the plane crashes is therefore:
$$p(text{crash})= p(text{object}) cdot p(text{crash | object}) + p(text{no object}) cdot p(text{crash | no object})$$
$$p(text{crash}) = frac{1}{10^6} cdot frac{1}{10^3} + 0 = frac{1}{10^9}$$
This is the probabiliy that the plane will crash every minute.
The total probabiliy that the plane will crash over an hour is:
$$p(text{hour crash}) = 1 - left( frac{10^9 - 1}{10^9} right)^{60} = 6 cdot 10^{-8} $$
The year has $365 cdot 24 = 8760$ hours.
And so, the probabiliy that the plane does crash is:
$$p(text{year crash})= 1 - (1 -6 cdot 10^{-8})^{8760}= 0.00053$$
Do you think my solution is correct?
probability discrete-mathematics
probability discrete-mathematics
asked Nov 28 '18 at 11:06
Aemilius
1,686314
asked Nov 28 '18 at 11:06
Aemilius
1,686314
asked Nov 28 '18 at 11:06
Aemilius
1,686314
asked Nov 28 '18 at 11:06
Aemilius
1,686314
asked Nov 28 '18 at 11:06
Aemilius
1,686314
1,686314
add a comment |
add a comment |
1 Answer
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Your solution is only an approximation (considering yearly probability), as $6cdot10^{-8} simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60cdot24cdot365=525600$ minutes. And finally the probability
begin{equation}
P(X)=1-bigg(frac{10^9-1}{10^9}bigg)^{525600}.
end{equation}
answered Nov 28 '18 at 13:09
vermator
795
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
add a comment |
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1 Answer
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1 Answer
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Your solution is only an approximation (considering yearly probability), as $6cdot10^{-8} simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60cdot24cdot365=525600$ minutes. And finally the probability
begin{equation}
P(X)=1-bigg(frac{10^9-1}{10^9}bigg)^{525600}.
end{equation}
answered Nov 28 '18 at 13:09
vermator
795
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
add a comment |
Your solution is only an approximation (considering yearly probability), as $6cdot10^{-8} simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60cdot24cdot365=525600$ minutes. And finally the probability
begin{equation}
P(X)=1-bigg(frac{10^9-1}{10^9}bigg)^{525600}.
end{equation}
answered Nov 28 '18 at 13:09
vermator
795
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
add a comment |
Your solution is only an approximation (considering yearly probability), as $6cdot10^{-8} simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60cdot24cdot365=525600$ minutes. And finally the probability
begin{equation}
P(X)=1-bigg(frac{10^9-1}{10^9}bigg)^{525600}.
end{equation}
answered Nov 28 '18 at 13:09
vermator
795
Your solution is only an approximation (considering yearly probability), as $6cdot10^{-8} simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60cdot24cdot365=525600$ minutes. And finally the probability
begin{equation}
P(X)=1-bigg(frac{10^9-1}{10^9}bigg)^{525600}.
end{equation}
answered Nov 28 '18 at 13:09
vermator
795
edited Nov 29 '18 at 22:15
answered Nov 28 '18 at 13:09
vermator
795
answered Nov 28 '18 at 13:09
vermator
795
answered Nov 28 '18 at 13:09
vermator
795
795
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
add a comment |
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
Does it mean that the fact that the runway is inspected does not change the overall probability?
– Aemilius
Nov 28 '18 at 13:16
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
IMO in current state of the problem yes. If the probability $r$ would be formulated "This object will cause a take-off disaster if not taken by a controller with probability $r$", then inspections would matter (but in that case we would need how often planes take-off).
– vermator
Nov 28 '18 at 13:29
add a comment |
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