Exponential distribution of decay
A container contains 13 particles, at time $t = 0$. The particles decay independently of each other and the time (unit: minutes) for a given particle's decay is a exponentially distributed random variable with expectation value $36.4$. Let $T$ denote the time that has passed when the number of particles has been reduced to 12. Calculate the probability $P(T > 1.035714)$
My solution is simple, but wrong. The pdf for the exponential distribution is given by
$$
f(x) = frac{1}{beta}exp(-x/beta), beta > 0, 0 < x < infty
$$
Numerically, we have
$$
f(x) = frac{1}{36.4}exp(-x/36.4)
$$
We want the probability
$$
P(T > 1.035714) = 1 - F(1.035714) = 1 - int_{0}^{1.035714}frac{1}{36.4}exp(-x/36.4)dx = exp{-1.035714/36.4} = 0.97194731
$$
I know this is the wrong answer, but I can't see where I go wrong.
exponential-distribution
add a comment |
A container contains 13 particles, at time $t = 0$. The particles decay independently of each other and the time (unit: minutes) for a given particle's decay is a exponentially distributed random variable with expectation value $36.4$. Let $T$ denote the time that has passed when the number of particles has been reduced to 12. Calculate the probability $P(T > 1.035714)$
My solution is simple, but wrong. The pdf for the exponential distribution is given by
$$
f(x) = frac{1}{beta}exp(-x/beta), beta > 0, 0 < x < infty
$$
Numerically, we have
$$
f(x) = frac{1}{36.4}exp(-x/36.4)
$$
We want the probability
$$
P(T > 1.035714) = 1 - F(1.035714) = 1 - int_{0}^{1.035714}frac{1}{36.4}exp(-x/36.4)dx = exp{-1.035714/36.4} = 0.97194731
$$
I know this is the wrong answer, but I can't see where I go wrong.
exponential-distribution
add a comment |
A container contains 13 particles, at time $t = 0$. The particles decay independently of each other and the time (unit: minutes) for a given particle's decay is a exponentially distributed random variable with expectation value $36.4$. Let $T$ denote the time that has passed when the number of particles has been reduced to 12. Calculate the probability $P(T > 1.035714)$
My solution is simple, but wrong. The pdf for the exponential distribution is given by
$$
f(x) = frac{1}{beta}exp(-x/beta), beta > 0, 0 < x < infty
$$
Numerically, we have
$$
f(x) = frac{1}{36.4}exp(-x/36.4)
$$
We want the probability
$$
P(T > 1.035714) = 1 - F(1.035714) = 1 - int_{0}^{1.035714}frac{1}{36.4}exp(-x/36.4)dx = exp{-1.035714/36.4} = 0.97194731
$$
I know this is the wrong answer, but I can't see where I go wrong.
exponential-distribution
A container contains 13 particles, at time $t = 0$. The particles decay independently of each other and the time (unit: minutes) for a given particle's decay is a exponentially distributed random variable with expectation value $36.4$. Let $T$ denote the time that has passed when the number of particles has been reduced to 12. Calculate the probability $P(T > 1.035714)$
My solution is simple, but wrong. The pdf for the exponential distribution is given by
$$
f(x) = frac{1}{beta}exp(-x/beta), beta > 0, 0 < x < infty
$$
Numerically, we have
$$
f(x) = frac{1}{36.4}exp(-x/36.4)
$$
We want the probability
$$
P(T > 1.035714) = 1 - F(1.035714) = 1 - int_{0}^{1.035714}frac{1}{36.4}exp(-x/36.4)dx = exp{-1.035714/36.4} = 0.97194731
$$
I know this is the wrong answer, but I can't see where I go wrong.
exponential-distribution
exponential-distribution
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
316
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1 Answer
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Hint:
That is the probability that a particular particle does not decay in that time
But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time
You know that their decays are independent of each other
Added after comments
There are at least two approaches giving the same answer:
If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} approx 0.6908$
The rate of decay for one particle is $frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
Hint:
That is the probability that a particular particle does not decay in that time
But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time
You know that their decays are independent of each other
Added after comments
There are at least two approaches giving the same answer:
If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} approx 0.6908$
The rate of decay for one particle is $frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
add a comment |
Hint:
That is the probability that a particular particle does not decay in that time
But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time
You know that their decays are independent of each other
Added after comments
There are at least two approaches giving the same answer:
If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} approx 0.6908$
The rate of decay for one particle is $frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
add a comment |
Hint:
That is the probability that a particular particle does not decay in that time
But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time
You know that their decays are independent of each other
Added after comments
There are at least two approaches giving the same answer:
If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} approx 0.6908$
The rate of decay for one particle is $frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $
Hint:
That is the probability that a particular particle does not decay in that time
But there are $13$ particles, (a clue is that $13$ does not appear in your calculation) and you want the probability that none of the $13$ decay in that time
You know that their decays are independent of each other
Added after comments
There are at least two approaches giving the same answer:
If the probability a particular particle does not decay in that time is about $0.97194731$, the the probability that none of the $13$ independently decay is about $0.97194731^{13} approx 0.6908$
The rate of decay for one particle is $frac{1}{36.4}$ so the rate for the first decay of $13$ independent particles is $frac{13}{36.4}$, making the probability that this does not happen by time $1.035714$ be $expleft(−1.035714 times frac{13}{36.4}right) approx 0.6908 $
edited Nov 26 '18 at 21:12
answered Nov 26 '18 at 19:43
Henry
98.1k475161
98.1k475161
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
add a comment |
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
Yeah. But where should I put $13$, and where should I put $12$. From the Physics I always learnt that the number of particles should be put as a coefficient before the exponential. Like $$ 12exp(-x/36.4), or 12exp(-x/36.4) $$ But according to the definition of the exponential distribution the exponent, and the coefficient should be the same. I'm confused.
– Kristoffer Jerzy Linder
Nov 26 '18 at 20:34
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
@KristofferJerzyLinder - multiplying your answer by $12$ will give a value over $1$ so it cannot be a probability. I have added a couple of approaches to my answer
– Henry
Nov 26 '18 at 21:13
add a comment |
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