Binominal distribution OR
$begingroup$
I got this exercise:
In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that:
she gets exactly 3 or exactly 4 questions right?
Should I understand this question as getting the EXACT value of 3 and EXACT value of 4, and then plus them together? Or is the question just calculating the EXACT for both 3 and 4? Right now I have just done a normal binomial calculation for exactly 3.
probability statistics binomial-distribution
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
$endgroup$
add a comment |
$begingroup$
I got this exercise:
In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that:
she gets exactly 3 or exactly 4 questions right?
Should I understand this question as getting the EXACT value of 3 and EXACT value of 4, and then plus them together? Or is the question just calculating the EXACT for both 3 and 4? Right now I have just done a normal binomial calculation for exactly 3.
probability statistics binomial-distribution
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
$endgroup$
1
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06
add a comment |
$begingroup$
I got this exercise:
In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that:
she gets exactly 3 or exactly 4 questions right?
Should I understand this question as getting the EXACT value of 3 and EXACT value of 4, and then plus them together? Or is the question just calculating the EXACT for both 3 and 4? Right now I have just done a normal binomial calculation for exactly 3.
probability statistics binomial-distribution
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
$endgroup$
I got this exercise:
In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that:
she gets exactly 3 or exactly 4 questions right?
Should I understand this question as getting the EXACT value of 3 and EXACT value of 4, and then plus them together? Or is the question just calculating the EXACT for both 3 and 4? Right now I have just done a normal binomial calculation for exactly 3.
probability statistics binomial-distribution
probability statistics binomial-distribution
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
asked Dec 1 '18 at 17:04
StudentCoderJavaStudentCoderJava
728
728
1
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06
add a comment |
1
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06
1
1
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The number of questions correct is $X sim mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5; pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)
x pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879 # P(X=3)
[5,] 4 0.0146 # P(X=4)
[6,] 5 0.0010
My interpretation of the English is that two separate
probabilities are requested. If asking for the sum,
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n choose k}p^k(1-p)^{n-k},$ for $k = 0, dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
$endgroup$
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
add a comment |
$begingroup$
You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 text{or} X = 4) = P(X=3) + P(X=4).$$
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
$endgroup$
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%2f3021571%2fbinominal-distribution-or%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 number of questions correct is $X sim mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5; pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)
x pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879 # P(X=3)
[5,] 4 0.0146 # P(X=4)
[6,] 5 0.0010
My interpretation of the English is that two separate
probabilities are requested. If asking for the sum,
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n choose k}p^k(1-p)^{n-k},$ for $k = 0, dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
$endgroup$
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
add a comment |
$begingroup$
The number of questions correct is $X sim mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5; pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)
x pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879 # P(X=3)
[5,] 4 0.0146 # P(X=4)
[6,] 5 0.0010
My interpretation of the English is that two separate
probabilities are requested. If asking for the sum,
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n choose k}p^k(1-p)^{n-k},$ for $k = 0, dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
$endgroup$
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
add a comment |
$begingroup$
The number of questions correct is $X sim mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5; pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)
x pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879 # P(X=3)
[5,] 4 0.0146 # P(X=4)
[6,] 5 0.0010
My interpretation of the English is that two separate
probabilities are requested. If asking for the sum,
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n choose k}p^k(1-p)^{n-k},$ for $k = 0, dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
$endgroup$
The number of questions correct is $X sim mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5; pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)
x pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879 # P(X=3)
[5,] 4 0.0146 # P(X=4)
[6,] 5 0.0010
My interpretation of the English is that two separate
probabilities are requested. If asking for the sum,
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n choose k}p^k(1-p)^{n-k},$ for $k = 0, dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
edited Dec 1 '18 at 23:42
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
answered Dec 1 '18 at 23:07
BruceETBruceET
35.3k71440
35.3k71440
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
add a comment |
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
$begingroup$
To be fair, the question is genuinely ambiguous--perhaps either written by a non-native speaker of English or by a native speaker goaded by non-mathematical copy editor say this. To ask clearly for two probabilties, it should be something like "... probability of exactly 3 questions right and probability of exactly 4 right."
$endgroup$
– BruceET
Dec 1 '18 at 23:28
add a comment |
$begingroup$
You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 text{or} X = 4) = P(X=3) + P(X=4).$$
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
$endgroup$
add a comment |
$begingroup$
You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 text{or} X = 4) = P(X=3) + P(X=4).$$
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
$endgroup$
add a comment |
$begingroup$
You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 text{or} X = 4) = P(X=3) + P(X=4).$$
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
$endgroup$
You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 text{or} X = 4) = P(X=3) + P(X=4).$$
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
answered Dec 1 '18 at 17:12
Sean RobersonSean Roberson
6,39031327
6,39031327
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%2f3021571%2fbinominal-distribution-or%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
1
$begingroup$
Well, you'll probably need to compute the exactly $3$ and exactly $4$ cases separately but then, yes. It looks like they want you to add them.
$endgroup$
– lulu
Dec 1 '18 at 17:06