Mixing
A caveau can be flooded for two reasons: slip or activation of fire extinguisher
$begingroup$
A caveau can be flooded for two reasons: slip or activation of fire extinguisher. If during the night no slip occurred, then for some reasons the fire extinguisher activate themself with probability $theta$. The split occurs with probability $0.01$
I want to find the probability that the slip occurred given that the caveau is flooded.
This is how I did it:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = 0.01(1-theta) $$
This is just my intuition: the slip occurred but the fire extinguishers dint go off.
However I tried doing it using Bayes theorem, but I find a different result:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = $$
$$ = frac{mathbb{P}[text{"Slip occured"}cap text{"Caveau flooded"}]}{mathbb{P}[text{"Caveau flooded"}]} = $$
$$ = frac{0.01}{0.01 + theta}$$
Can you help me finding my mistake?
probability
asked Jan 15 at 18:37
qcc101qcc101
622213
$endgroup$
add a comment |
$begingroup$
A caveau can be flooded for two reasons: slip or activation of fire extinguisher. If during the night no slip occurred, then for some reasons the fire extinguisher activate themself with probability $theta$. The split occurs with probability $0.01$
I want to find the probability that the slip occurred given that the caveau is flooded.
This is how I did it:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = 0.01(1-theta) $$
This is just my intuition: the slip occurred but the fire extinguishers dint go off.
However I tried doing it using Bayes theorem, but I find a different result:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = $$
$$ = frac{mathbb{P}[text{"Slip occured"}cap text{"Caveau flooded"}]}{mathbb{P}[text{"Caveau flooded"}]} = $$
$$ = frac{0.01}{0.01 + theta}$$
Can you help me finding my mistake?
probability
asked Jan 15 at 18:37
qcc101qcc101
622213
$endgroup$
add a comment |
$begingroup$
A caveau can be flooded for two reasons: slip or activation of fire extinguisher. If during the night no slip occurred, then for some reasons the fire extinguisher activate themself with probability $theta$. The split occurs with probability $0.01$
I want to find the probability that the slip occurred given that the caveau is flooded.
This is how I did it:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = 0.01(1-theta) $$
This is just my intuition: the slip occurred but the fire extinguishers dint go off.
However I tried doing it using Bayes theorem, but I find a different result:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = $$
$$ = frac{mathbb{P}[text{"Slip occured"}cap text{"Caveau flooded"}]}{mathbb{P}[text{"Caveau flooded"}]} = $$
$$ = frac{0.01}{0.01 + theta}$$
Can you help me finding my mistake?
probability
asked Jan 15 at 18:37
qcc101qcc101
622213
$endgroup$
A caveau can be flooded for two reasons: slip or activation of fire extinguisher. If during the night no slip occurred, then for some reasons the fire extinguisher activate themself with probability $theta$. The split occurs with probability $0.01$
I want to find the probability that the slip occurred given that the caveau is flooded.
This is how I did it:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = 0.01(1-theta) $$
This is just my intuition: the slip occurred but the fire extinguishers dint go off.
However I tried doing it using Bayes theorem, but I find a different result:
$$ mathbb{P}[text{"Slip occured"} vert text{"Caveau flooded"}] = $$
$$ = frac{mathbb{P}[text{"Slip occured"}cap text{"Caveau flooded"}]}{mathbb{P}[text{"Caveau flooded"}]} = $$
$$ = frac{0.01}{0.01 + theta}$$
Can you help me finding my mistake?
probability
probability
asked Jan 15 at 18:37
qcc101qcc101
622213
asked Jan 15 at 18:37
qcc101qcc101
622213
asked Jan 15 at 18:37
qcc101qcc101
622213
asked Jan 15 at 18:37
qcc101qcc101
622213
asked Jan 15 at 18:37
qcc101qcc101
622213
622213
add a comment |
add a comment |
1 Answer
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$begingroup$
Your mistake lies in the calculation of the total probability of caveau flooding. There are two ways the caveau can be flooded. Either a slip occurs or no slip occurs and the fire extinguishers start themselves. So
$$E_1 - text{A slip occurs}$$
$$E_2 - text{No slip occurs and the fire extinguishers start themselves}$$
As you correctly calculated
$$P(E_1) = 0.01$$
But the second one will be
$$P(E_2) = (1-0.01)theta$$
First terms is as no slip occurs and the second is the probability that fire extinguishers will autostart.
So the probability that slip occurred given that caveau is flooded (using Bayes theorem)
$$P= frac{0.01}{0.01 + 0.99theta}$$
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Your mistake lies in the calculation of the total probability of caveau flooding. There are two ways the caveau can be flooded. Either a slip occurs or no slip occurs and the fire extinguishers start themselves. So
$$E_1 - text{A slip occurs}$$
$$E_2 - text{No slip occurs and the fire extinguishers start themselves}$$
As you correctly calculated
$$P(E_1) = 0.01$$
But the second one will be
$$P(E_2) = (1-0.01)theta$$
First terms is as no slip occurs and the second is the probability that fire extinguishers will autostart.
So the probability that slip occurred given that caveau is flooded (using Bayes theorem)
$$P= frac{0.01}{0.01 + 0.99theta}$$
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
add a comment |
$begingroup$
Your mistake lies in the calculation of the total probability of caveau flooding. There are two ways the caveau can be flooded. Either a slip occurs or no slip occurs and the fire extinguishers start themselves. So
$$E_1 - text{A slip occurs}$$
$$E_2 - text{No slip occurs and the fire extinguishers start themselves}$$
As you correctly calculated
$$P(E_1) = 0.01$$
But the second one will be
$$P(E_2) = (1-0.01)theta$$
First terms is as no slip occurs and the second is the probability that fire extinguishers will autostart.
So the probability that slip occurred given that caveau is flooded (using Bayes theorem)
$$P= frac{0.01}{0.01 + 0.99theta}$$
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
$endgroup$
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
add a comment |
$begingroup$
Your mistake lies in the calculation of the total probability of caveau flooding. There are two ways the caveau can be flooded. Either a slip occurs or no slip occurs and the fire extinguishers start themselves. So
$$E_1 - text{A slip occurs}$$
$$E_2 - text{No slip occurs and the fire extinguishers start themselves}$$
As you correctly calculated
$$P(E_1) = 0.01$$
But the second one will be
$$P(E_2) = (1-0.01)theta$$
First terms is as no slip occurs and the second is the probability that fire extinguishers will autostart.
So the probability that slip occurred given that caveau is flooded (using Bayes theorem)
$$P= frac{0.01}{0.01 + 0.99theta}$$
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
$endgroup$
Your mistake lies in the calculation of the total probability of caveau flooding. There are two ways the caveau can be flooded. Either a slip occurs or no slip occurs and the fire extinguishers start themselves. So
$$E_1 - text{A slip occurs}$$
$$E_2 - text{No slip occurs and the fire extinguishers start themselves}$$
As you correctly calculated
$$P(E_1) = 0.01$$
But the second one will be
$$P(E_2) = (1-0.01)theta$$
First terms is as no slip occurs and the second is the probability that fire extinguishers will autostart.
So the probability that slip occurred given that caveau is flooded (using Bayes theorem)
$$P= frac{0.01}{0.01 + 0.99theta}$$
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
answered Jan 15 at 18:56
Sauhard SharmaSauhard Sharma
953318
953318
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
add a comment |
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
Thank you! I have a follow up question: I need to build a confidence interval of 95% for the probability that slip occurred without knowing that there is flood. I also now know that $theta = 0.53$ I need to use Wald standard interval, which is: $$ overline{X} pm Z_{frac{alpha}{2}} frac{sigma}{sqrt{n}}$$ However what are the $overline{X}$ and $sigma$ in this case?
$endgroup$
– qcc101
Jan 15 at 19:06
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
@qcc101 - You'd need random sampling and based on that you will be estimating the parameter(probability that slip occurred). The terms $bar{X}$ and $sigma$ come from there. Also, this is an entirely separate question in itself. As is the policy of MSE, please ask the question separately.
$endgroup$
– Sauhard Sharma
Jan 15 at 19:21
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
$begingroup$
I see, I will try to work it out myself with your hint and then post again If I don't succeed. Thank you!
$endgroup$
– qcc101
Jan 15 at 19:26
add a comment |
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