Whether to use conditional probability or Bayes theorem?
$begingroup$
I had found this question.
A group of boys and girls know either French or Spanish. The number of boys and girls are in the ratio $1:4$. $30%$ of the girls know Spanish and the rest of them know French. On the other hand, $50%$ of the boys know Spanish and the rest of them know French. A student is chosen at random from the group of students who know Spanish. What is the probability that the chosen student is a girl?
My attempt was:
Let $P(G)$ and $P(B)$ be probability of choosing a girl and boy respectively. And $P(S)$ be probability of choosing someone who knows Spanish. Then,
$P(G|S)= frac{P(S|G)*P(G)}{P(S|G)*P(G)+P(S|B)*P(B)}$
But the answer using this way is not in the option. Moreover (B) is the answer in their answer key.
(A) 2/7
(B) 12/17
(C) 20/41
(D) 8/13
Is something wrong with my approach? How can I approach it correctly?
probability
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
$endgroup$
add a comment |
$begingroup$
I had found this question.
A group of boys and girls know either French or Spanish. The number of boys and girls are in the ratio $1:4$. $30%$ of the girls know Spanish and the rest of them know French. On the other hand, $50%$ of the boys know Spanish and the rest of them know French. A student is chosen at random from the group of students who know Spanish. What is the probability that the chosen student is a girl?
My attempt was:
Let $P(G)$ and $P(B)$ be probability of choosing a girl and boy respectively. And $P(S)$ be probability of choosing someone who knows Spanish. Then,
$P(G|S)= frac{P(S|G)*P(G)}{P(S|G)*P(G)+P(S|B)*P(B)}$
But the answer using this way is not in the option. Moreover (B) is the answer in their answer key.
(A) 2/7
(B) 12/17
(C) 20/41
(D) 8/13
Is something wrong with my approach? How can I approach it correctly?
probability
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
$endgroup$
add a comment |
$begingroup$
I had found this question.
A group of boys and girls know either French or Spanish. The number of boys and girls are in the ratio $1:4$. $30%$ of the girls know Spanish and the rest of them know French. On the other hand, $50%$ of the boys know Spanish and the rest of them know French. A student is chosen at random from the group of students who know Spanish. What is the probability that the chosen student is a girl?
My attempt was:
Let $P(G)$ and $P(B)$ be probability of choosing a girl and boy respectively. And $P(S)$ be probability of choosing someone who knows Spanish. Then,
$P(G|S)= frac{P(S|G)*P(G)}{P(S|G)*P(G)+P(S|B)*P(B)}$
But the answer using this way is not in the option. Moreover (B) is the answer in their answer key.
(A) 2/7
(B) 12/17
(C) 20/41
(D) 8/13
Is something wrong with my approach? How can I approach it correctly?
probability
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
$endgroup$
I had found this question.
A group of boys and girls know either French or Spanish. The number of boys and girls are in the ratio $1:4$. $30%$ of the girls know Spanish and the rest of them know French. On the other hand, $50%$ of the boys know Spanish and the rest of them know French. A student is chosen at random from the group of students who know Spanish. What is the probability that the chosen student is a girl?
My attempt was:
Let $P(G)$ and $P(B)$ be probability of choosing a girl and boy respectively. And $P(S)$ be probability of choosing someone who knows Spanish. Then,
$P(G|S)= frac{P(S|G)*P(G)}{P(S|G)*P(G)+P(S|B)*P(B)}$
But the answer using this way is not in the option. Moreover (B) is the answer in their answer key.
(A) 2/7
(B) 12/17
(C) 20/41
(D) 8/13
Is something wrong with my approach? How can I approach it correctly?
probability
probability
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
edited Dec 23 '18 at 10:17
Rebellos
15.1k31250
15.1k31250
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
asked Dec 23 '18 at 10:15
Mr.Sigma.Mr.Sigma.
18310
18310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probability of choosing a girl is $frac{4}{5}$ probability that a girl speaks Spanish is $frac{4}{5}.frac{3}{10}=frac{12}{50}$ similarly probability of selecting a boy speaking Spanish is $frac{1}{10}$ Thus using Bayes formula we get the answer as $frac{12}{17}$
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
$endgroup$
add a comment |
$begingroup$
It is IMV handsome to choose for $20$ boys and $80$ girls.
Then $24$ girls speak Spanish and $56$ girls speak French.
And $10$ boys speak Spanish and $10$ boys speak French.
So if a Spanish speaking student is chosen then the probability that it is a girl equals:$$frac{24}{24+10}=frac{24}{34}=frac{12}{17}$$
Using your method I find:
$$Pleft(Gmid Sright)Pleft(Sright)=Pleft(Smid Gright)Pleft(Gright)$$
with:
$Pleft(Smid Gright)=frac{3}{10}$- $Pleft(Gright)=frac{4}{5}$
- $Pleft(Sright)=Pleft(Smid Bright)Pleft(Bright)+Pleft(Smid Gright)Pleft(Gright)=frac{1}{2}frac{1}{5}+frac{3}{10}frac{4}{5}=frac{17}{50}$
and again leading to $Pleft(Gmid Sright)=frac{12}{17}$.
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
$endgroup$
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
});
}
});
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%2f3050221%2fwhether-to-use-conditional-probability-or-bayes-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability of choosing a girl is $frac{4}{5}$ probability that a girl speaks Spanish is $frac{4}{5}.frac{3}{10}=frac{12}{50}$ similarly probability of selecting a boy speaking Spanish is $frac{1}{10}$ Thus using Bayes formula we get the answer as $frac{12}{17}$
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
$endgroup$
add a comment |
$begingroup$
The probability of choosing a girl is $frac{4}{5}$ probability that a girl speaks Spanish is $frac{4}{5}.frac{3}{10}=frac{12}{50}$ similarly probability of selecting a boy speaking Spanish is $frac{1}{10}$ Thus using Bayes formula we get the answer as $frac{12}{17}$
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
$endgroup$
add a comment |
$begingroup$
The probability of choosing a girl is $frac{4}{5}$ probability that a girl speaks Spanish is $frac{4}{5}.frac{3}{10}=frac{12}{50}$ similarly probability of selecting a boy speaking Spanish is $frac{1}{10}$ Thus using Bayes formula we get the answer as $frac{12}{17}$
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
$endgroup$
The probability of choosing a girl is $frac{4}{5}$ probability that a girl speaks Spanish is $frac{4}{5}.frac{3}{10}=frac{12}{50}$ similarly probability of selecting a boy speaking Spanish is $frac{1}{10}$ Thus using Bayes formula we get the answer as $frac{12}{17}$
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
answered Dec 23 '18 at 10:27
Archis WelankarArchis Welankar
12.1k41642
12.1k41642
add a comment |
add a comment |
$begingroup$
It is IMV handsome to choose for $20$ boys and $80$ girls.
Then $24$ girls speak Spanish and $56$ girls speak French.
And $10$ boys speak Spanish and $10$ boys speak French.
So if a Spanish speaking student is chosen then the probability that it is a girl equals:$$frac{24}{24+10}=frac{24}{34}=frac{12}{17}$$
Using your method I find:
$$Pleft(Gmid Sright)Pleft(Sright)=Pleft(Smid Gright)Pleft(Gright)$$
with:
$Pleft(Smid Gright)=frac{3}{10}$- $Pleft(Gright)=frac{4}{5}$
- $Pleft(Sright)=Pleft(Smid Bright)Pleft(Bright)+Pleft(Smid Gright)Pleft(Gright)=frac{1}{2}frac{1}{5}+frac{3}{10}frac{4}{5}=frac{17}{50}$
and again leading to $Pleft(Gmid Sright)=frac{12}{17}$.
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is IMV handsome to choose for $20$ boys and $80$ girls.
Then $24$ girls speak Spanish and $56$ girls speak French.
And $10$ boys speak Spanish and $10$ boys speak French.
So if a Spanish speaking student is chosen then the probability that it is a girl equals:$$frac{24}{24+10}=frac{24}{34}=frac{12}{17}$$
Using your method I find:
$$Pleft(Gmid Sright)Pleft(Sright)=Pleft(Smid Gright)Pleft(Gright)$$
with:
$Pleft(Smid Gright)=frac{3}{10}$- $Pleft(Gright)=frac{4}{5}$
- $Pleft(Sright)=Pleft(Smid Bright)Pleft(Bright)+Pleft(Smid Gright)Pleft(Gright)=frac{1}{2}frac{1}{5}+frac{3}{10}frac{4}{5}=frac{17}{50}$
and again leading to $Pleft(Gmid Sright)=frac{12}{17}$.
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is IMV handsome to choose for $20$ boys and $80$ girls.
Then $24$ girls speak Spanish and $56$ girls speak French.
And $10$ boys speak Spanish and $10$ boys speak French.
So if a Spanish speaking student is chosen then the probability that it is a girl equals:$$frac{24}{24+10}=frac{24}{34}=frac{12}{17}$$
Using your method I find:
$$Pleft(Gmid Sright)Pleft(Sright)=Pleft(Smid Gright)Pleft(Gright)$$
with:
$Pleft(Smid Gright)=frac{3}{10}$- $Pleft(Gright)=frac{4}{5}$
- $Pleft(Sright)=Pleft(Smid Bright)Pleft(Bright)+Pleft(Smid Gright)Pleft(Gright)=frac{1}{2}frac{1}{5}+frac{3}{10}frac{4}{5}=frac{17}{50}$
and again leading to $Pleft(Gmid Sright)=frac{12}{17}$.
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
$endgroup$
It is IMV handsome to choose for $20$ boys and $80$ girls.
Then $24$ girls speak Spanish and $56$ girls speak French.
And $10$ boys speak Spanish and $10$ boys speak French.
So if a Spanish speaking student is chosen then the probability that it is a girl equals:$$frac{24}{24+10}=frac{24}{34}=frac{12}{17}$$
Using your method I find:
$$Pleft(Gmid Sright)Pleft(Sright)=Pleft(Smid Gright)Pleft(Gright)$$
with:
$Pleft(Smid Gright)=frac{3}{10}$- $Pleft(Gright)=frac{4}{5}$
- $Pleft(Sright)=Pleft(Smid Bright)Pleft(Bright)+Pleft(Smid Gright)Pleft(Gright)=frac{1}{2}frac{1}{5}+frac{3}{10}frac{4}{5}=frac{17}{50}$
and again leading to $Pleft(Gmid Sright)=frac{12}{17}$.
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
edited Dec 23 '18 at 10:45
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
answered Dec 23 '18 at 10:35
drhabdrhab
102k545136
102k545136
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.
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%2f3050221%2fwhether-to-use-conditional-probability-or-bayes-theorem%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
