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
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
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
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
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
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
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
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
asked Nov 26 '18 at 19:34
Kristoffer Jerzy Linder
316
316
add a comment |
add a comment |
1 Answer
1
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 $
answered Nov 26 '18 at 19:43
Henry
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014805%2fexponential-distribution-of-decay%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 $
answered Nov 26 '18 at 19:43
Henry
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 |
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 $
answered Nov 26 '18 at 19:43
Henry
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 |
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 $
answered Nov 26 '18 at 19:43
Henry
98.1k475161
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 $
answered Nov 26 '18 at 19:43
Henry
98.1k475161
edited Nov 26 '18 at 21:12
answered Nov 26 '18 at 19:43
Henry
98.1k475161
answered Nov 26 '18 at 19:43
Henry
98.1k475161
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014805%2fexponential-distribution-of-decay%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
