Largest possible value of a probability
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
add a comment |
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
4
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32
add a comment |
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.
Try
Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,
$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$
notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,
$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$
$$ P(X=3) = 1.8-p-2p^2 $$
take derivative with respect to $p$ gives
$$ (P(X=3))' = -1 - 4p $$
which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?
probability
probability
edited Nov 28 at 19:51
LuxGiammi
16410
16410
asked Nov 28 at 18:37
Neymar
385113
385113
4
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32
add a comment |
4
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32
4
4
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
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%2f3017524%2flargest-possible-value-of-a-probability%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
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
add a comment |
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
add a comment |
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
If your trials are independent
$P(X=3)= 0.6^3$
But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.
All of the trials have equal likelihood of showing disease, but they are not independent.
If $q$ is big... i.e. let $q = 1$
Either all three samples show disease or none do.
$P(X=3) = 0.6$
At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.
$P(X=3) = 0$
edited Nov 28 at 19:48
answered Nov 28 at 19:38
Doug M
43.9k31854
43.9k31854
add a comment |
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%2f3017524%2flargest-possible-value-of-a-probability%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
4
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
– LuxGiammi
Nov 28 at 19:32