Is this an application case of Bayes theorem? Is my book wrong?Bayes theorem applicationAn explanation of how this solution is derivedGoing wrong with Bayes theoremApplication of Bayes-theoremBasic application of Bayes theoremRoot estimator to get variance?Is this application of Bayes' theorem correct?Problem based on law of total probability and bayes theoremBayes theorem in geneticsBayes' Theorem, what am I doing wrong?
Help! My Character is too much for her story!
How do I raise a figure (placed with wrapfig) to be flush with the top of a paragraph?
Cycles on the torus
I am the person who abides by rules, but breaks the rules. Who am I?
Trocar background-image com delay via jQuery
Is "cogitate" used appropriately in "I cogitate that success relies on hard work"?
Use Mercury as quenching liquid for swords?
Are E natural minor and B harmonic minor related?
Why does this boat have a landing pad? (SpaceX's GO Searcher) Any plans for propulsive capsule landings?
Strange opamp's output impedance in spice
PTIJ: Who was the sixth set of priestly clothes for?
Does an unused member variable take up memory?
How do spaceships determine each other's mass in space?
-1 to the power of a irrational number
Boss Telling direct supervisor I snitched
How do you make a gun that shoots melee weapons and/or swords?
"If + would" conditional in present perfect tense
What happened to the colonial estates belonging to loyalists after the American Revolution?
What does *dead* mean in *What do you mean, dead?*?
Can the Witch Sight warlock invocation see through the Mirror Image spell?
Has a sovereign Communist government ever run, and conceded loss, on a fair election?
Under what conditions can the right to be silence be revoked in the USA?
Why do we say 'Pairwise Disjoint', rather than 'Disjoint'?
Does the US political system, in principle, allow for a no-party system?
Is this an application case of Bayes theorem? Is my book wrong?
Bayes theorem applicationAn explanation of how this solution is derivedGoing wrong with Bayes theoremApplication of Bayes-theoremBasic application of Bayes theoremRoot estimator to get variance?Is this application of Bayes' theorem correct?Problem based on law of total probability and bayes theoremBayes theorem in geneticsBayes' Theorem, what am I doing wrong?
$begingroup$
A sickness has a heterozygote frequency of $frac120$ ie. 1 in 20 people of a population will have the allele combination $Aa$ where a denotes the recessive and A the dominant allele. To become sick the carrier needs to have two $aa$ alleles.
A mother is known to have the combination $Aa$ , what is the probability her child will be sick (have $aa$) if the combination of the husband is unknown (i.e. $AA, Aa, aa)$?
Now here is where I struggle. If the husband has $AA$ , the child will never get $aa$, if the husband has $Aa$, the child will get $aa$ with a chance of $frac14$.
If the husband has $aa$ (is sick himself) then the child wil get $aa$ with a chance of $frac12$
The book states that the answer a priori is for the child to be sick in this situation is $frac120 frac12 frac12 = frac180$
Shouldn't it instead be : $frac14frac120 + (textbfdouble aa frequency)cdot frac12$?
statistics bayes-theorem
edited yesterday
rash
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 6 more comments
$begingroup$
A sickness has a heterozygote frequency of $frac120$ ie. 1 in 20 people of a population will have the allele combination $Aa$ where a denotes the recessive and A the dominant allele. To become sick the carrier needs to have two $aa$ alleles.
A mother is known to have the combination $Aa$ , what is the probability her child will be sick (have $aa$) if the combination of the husband is unknown (i.e. $AA, Aa, aa)$?
Now here is where I struggle. If the husband has $AA$ , the child will never get $aa$, if the husband has $Aa$, the child will get $aa$ with a chance of $frac14$.
If the husband has $aa$ (is sick himself) then the child wil get $aa$ with a chance of $frac12$
The book states that the answer a priori is for the child to be sick in this situation is $frac120 frac12 frac12 = frac180$
Shouldn't it instead be : $frac14frac120 + (textbfdouble aa frequency)cdot frac12$?
statistics bayes-theorem
edited yesterday
rash
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday
|
show 6 more comments
$begingroup$
A sickness has a heterozygote frequency of $frac120$ ie. 1 in 20 people of a population will have the allele combination $Aa$ where a denotes the recessive and A the dominant allele. To become sick the carrier needs to have two $aa$ alleles.
A mother is known to have the combination $Aa$ , what is the probability her child will be sick (have $aa$) if the combination of the husband is unknown (i.e. $AA, Aa, aa)$?
Now here is where I struggle. If the husband has $AA$ , the child will never get $aa$, if the husband has $Aa$, the child will get $aa$ with a chance of $frac14$.
If the husband has $aa$ (is sick himself) then the child wil get $aa$ with a chance of $frac12$
The book states that the answer a priori is for the child to be sick in this situation is $frac120 frac12 frac12 = frac180$
Shouldn't it instead be : $frac14frac120 + (textbfdouble aa frequency)cdot frac12$?
statistics bayes-theorem
edited yesterday
rash
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A sickness has a heterozygote frequency of $frac120$ ie. 1 in 20 people of a population will have the allele combination $Aa$ where a denotes the recessive and A the dominant allele. To become sick the carrier needs to have two $aa$ alleles.
A mother is known to have the combination $Aa$ , what is the probability her child will be sick (have $aa$) if the combination of the husband is unknown (i.e. $AA, Aa, aa)$?
Now here is where I struggle. If the husband has $AA$ , the child will never get $aa$, if the husband has $Aa$, the child will get $aa$ with a chance of $frac14$.
If the husband has $aa$ (is sick himself) then the child wil get $aa$ with a chance of $frac12$
The book states that the answer a priori is for the child to be sick in this situation is $frac120 frac12 frac12 = frac180$
Shouldn't it instead be : $frac14frac120 + (textbfdouble aa frequency)cdot frac12$?
statistics bayes-theorem
statistics bayes-theorem
edited yesterday
rash
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
rash
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
rash
20811
edited yesterday
rash
20811
edited yesterday
rash
20811
20811
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
asked yesterday
Jonathan MaximilianJonathan Maximilian
161
161
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jonathan Maximilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday
|
show 6 more comments
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
);
);
Jonathan Maximilian is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3140468%2fis-this-an-application-case-of-bayes-theorem-is-my-book-wrong%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
$begingroup$
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You have mentioned that the father could be AA, Aa, or aa. What we need to remember is that these represent the alleles he received from his parents. So, if we let the first letter represent the allele received from his father and the second letter represent the allele gained from his mother, there are actually four possibilities: AA, Aa, aA, aa. That is, the father has a 1/4 chance of being either AA or aa, but a 2/4 = 1/2 chance of being Aa.
When these are combined with the known Aa of the mother, you will see that there are 4 of 16 possibilities resulting in aa. Thus, the probability of being aa are 1/4 ... and if 1/20 of those have the disease, then the result is 1/80.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Den ZelleDen Zelle
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Den ZelleDen Zelle
113
answered yesterday
Den ZelleDen Zelle
113
113
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Den Zelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Jonathan Maximilian is a new contributor. Be nice, and check out our Code of Conduct.
Jonathan Maximilian is a new contributor. Be nice, and check out our Code of Conduct.
Jonathan Maximilian is a new contributor. Be nice, and check out our Code of Conduct.
Jonathan Maximilian is a new contributor. Be nice, and check out our Code of Conduct.
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.
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%2f3140468%2fis-this-an-application-case-of-bayes-theorem-is-my-book-wrong%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
$begingroup$
Maybe the implication is that if the husband's combination is unknown, it must mean that he does not have $aa$, since he would have shown symptoms already? Is the "(i.e., $AA, Aa, aa$)" yours, or the book's?
$endgroup$
– Brian Tung
yesterday
$begingroup$
mine, and your answer makes a lot of sense lol, but my assumption is correct if aa was included ?
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
actually i rechecked, and the book doesnt say anything about the husband showing early symptoms, it just says that the probability a priori is what i wrote in my post (1/80)
$endgroup$
– Jonathan Maximilian
yesterday
$begingroup$
Yeah, I don't know. Is the book online by any chance?
$endgroup$
– Brian Tung
yesterday
$begingroup$
no and its not in english lol, do you know if my attempt is correct? can you just sum the two probability trees together?
$endgroup$
– Jonathan Maximilian
yesterday