Mixing
![Damped & Negatively damped harmonic oscillator [not physics question]](https://lh3.googleusercontent.com/-mas791g22hc/AAAAAAAAAAI/AAAAAAAAEzU/eEbEspgWUMA/s72-c/photo.jpg?sz=32)
Probability without replacement with an urn with five balls labeled 1 through 5
$begingroup$
An urn contains five balls labeled 1 through $5.$ Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let $A$ be the event that the second ball drawn is ball $4$ or $5.$ Find $P(A).$
(ii) Let $B$ be the event that the sample of three balls contains ball $1$ or ball $2$ or both. Find $P(B).$
(iii) Let $C$ be the event that the three recorded numbers come in increasing order. Find $P(C).$
For part (i) I have that computed the probability to be $P(A)= largefrac{4}{5}cdotfrac{1}{4}+frac{4}{5}cdotfrac{1}{5}.$ I do this because we can decompose the event in question into the event in which ball $4$ is drawn in the second trial, and the event in which ball $5$ is drawn in the second trial, which are disjoint.
For part (ii) I have that we can decompose the event into three disjoint events, which are the simpler events listed. It follows that the probability of the event in question is equal to $2large left( frac{1}{5}cdotfrac{1}{4}cdotfrac{1}{3}right)+frac{2}{5}cdotfrac{2}{4}cdotfrac{2}{3}.$
For part (iii) I have that there are $11$ possible was to choose the numbers in increasing order. Then we have that the probability is $11$ divided by the total number of choosing $3$ balls out of $5$ without replacement.
Is the above reasoning correct? If not, could you please point out my mistakes?
probability probability-theory
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
add a comment |
$begingroup$
An urn contains five balls labeled 1 through $5.$ Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let $A$ be the event that the second ball drawn is ball $4$ or $5.$ Find $P(A).$
(ii) Let $B$ be the event that the sample of three balls contains ball $1$ or ball $2$ or both. Find $P(B).$
(iii) Let $C$ be the event that the three recorded numbers come in increasing order. Find $P(C).$
For part (i) I have that computed the probability to be $P(A)= largefrac{4}{5}cdotfrac{1}{4}+frac{4}{5}cdotfrac{1}{5}.$ I do this because we can decompose the event in question into the event in which ball $4$ is drawn in the second trial, and the event in which ball $5$ is drawn in the second trial, which are disjoint.
For part (ii) I have that we can decompose the event into three disjoint events, which are the simpler events listed. It follows that the probability of the event in question is equal to $2large left( frac{1}{5}cdotfrac{1}{4}cdotfrac{1}{3}right)+frac{2}{5}cdotfrac{2}{4}cdotfrac{2}{3}.$
For part (iii) I have that there are $11$ possible was to choose the numbers in increasing order. Then we have that the probability is $11$ divided by the total number of choosing $3$ balls out of $5$ without replacement.
Is the above reasoning correct? If not, could you please point out my mistakes?
probability probability-theory
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
1
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42
add a comment |
$begingroup$
An urn contains five balls labeled 1 through $5.$ Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let $A$ be the event that the second ball drawn is ball $4$ or $5.$ Find $P(A).$
(ii) Let $B$ be the event that the sample of three balls contains ball $1$ or ball $2$ or both. Find $P(B).$
(iii) Let $C$ be the event that the three recorded numbers come in increasing order. Find $P(C).$
For part (i) I have that computed the probability to be $P(A)= largefrac{4}{5}cdotfrac{1}{4}+frac{4}{5}cdotfrac{1}{5}.$ I do this because we can decompose the event in question into the event in which ball $4$ is drawn in the second trial, and the event in which ball $5$ is drawn in the second trial, which are disjoint.
For part (ii) I have that we can decompose the event into three disjoint events, which are the simpler events listed. It follows that the probability of the event in question is equal to $2large left( frac{1}{5}cdotfrac{1}{4}cdotfrac{1}{3}right)+frac{2}{5}cdotfrac{2}{4}cdotfrac{2}{3}.$
For part (iii) I have that there are $11$ possible was to choose the numbers in increasing order. Then we have that the probability is $11$ divided by the total number of choosing $3$ balls out of $5$ without replacement.
Is the above reasoning correct? If not, could you please point out my mistakes?
probability probability-theory
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
An urn contains five balls labeled 1 through $5.$ Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let $A$ be the event that the second ball drawn is ball $4$ or $5.$ Find $P(A).$
(ii) Let $B$ be the event that the sample of three balls contains ball $1$ or ball $2$ or both. Find $P(B).$
(iii) Let $C$ be the event that the three recorded numbers come in increasing order. Find $P(C).$
For part (i) I have that computed the probability to be $P(A)= largefrac{4}{5}cdotfrac{1}{4}+frac{4}{5}cdotfrac{1}{5}.$ I do this because we can decompose the event in question into the event in which ball $4$ is drawn in the second trial, and the event in which ball $5$ is drawn in the second trial, which are disjoint.
For part (ii) I have that we can decompose the event into three disjoint events, which are the simpler events listed. It follows that the probability of the event in question is equal to $2large left( frac{1}{5}cdotfrac{1}{4}cdotfrac{1}{3}right)+frac{2}{5}cdotfrac{2}{4}cdotfrac{2}{3}.$
For part (iii) I have that there are $11$ possible was to choose the numbers in increasing order. Then we have that the probability is $11$ divided by the total number of choosing $3$ balls out of $5$ without replacement.
Is the above reasoning correct? If not, could you please point out my mistakes?
probability probability-theory
probability probability-theory
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 11:37
Gaby AlfonsoGaby Alfonso
1,1951418
1,1951418
1
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42
add a comment |
1
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42
1
1
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since there is no information on the first draw, this probability simply equals:
$$P(A) = frac{2}{5}$$
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as:
$$P(A) = frac{2}{5} frac{1}{4} + frac{3}{5} frac{2}{4} = frac{8}{20} = frac{2}{5}$$The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
$$P(B) = 1 - frac{3}{5} frac{2}{4} frac{1}{3} = 1 - frac{1}{10} = frac{9}{10}$$There are ${5 choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find:
$$P(C) = frac{{5 choose 3}}{5 cdot 4 cdot 3} = frac{10}{60} = frac{1}{6}$$
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
$endgroup$
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%2f3096120%2fprobability-without-replacement-with-an-urn-with-five-balls-labeled-1-through-5%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
$begingroup$
Since there is no information on the first draw, this probability simply equals:
$$P(A) = frac{2}{5}$$
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as:
$$P(A) = frac{2}{5} frac{1}{4} + frac{3}{5} frac{2}{4} = frac{8}{20} = frac{2}{5}$$The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
$$P(B) = 1 - frac{3}{5} frac{2}{4} frac{1}{3} = 1 - frac{1}{10} = frac{9}{10}$$There are ${5 choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find:
$$P(C) = frac{{5 choose 3}}{5 cdot 4 cdot 3} = frac{10}{60} = frac{1}{6}$$
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
Since there is no information on the first draw, this probability simply equals:
$$P(A) = frac{2}{5}$$
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as:
$$P(A) = frac{2}{5} frac{1}{4} + frac{3}{5} frac{2}{4} = frac{8}{20} = frac{2}{5}$$The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
$$P(B) = 1 - frac{3}{5} frac{2}{4} frac{1}{3} = 1 - frac{1}{10} = frac{9}{10}$$There are ${5 choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find:
$$P(C) = frac{{5 choose 3}}{5 cdot 4 cdot 3} = frac{10}{60} = frac{1}{6}$$
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
$endgroup$
add a comment |
$begingroup$
Since there is no information on the first draw, this probability simply equals:
$$P(A) = frac{2}{5}$$
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as:
$$P(A) = frac{2}{5} frac{1}{4} + frac{3}{5} frac{2}{4} = frac{8}{20} = frac{2}{5}$$The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
$$P(B) = 1 - frac{3}{5} frac{2}{4} frac{1}{3} = 1 - frac{1}{10} = frac{9}{10}$$There are ${5 choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find:
$$P(C) = frac{{5 choose 3}}{5 cdot 4 cdot 3} = frac{10}{60} = frac{1}{6}$$
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
$endgroup$
Since there is no information on the first draw, this probability simply equals:
$$P(A) = frac{2}{5}$$
Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as:
$$P(A) = frac{2}{5} frac{1}{4} + frac{3}{5} frac{2}{4} = frac{8}{20} = frac{2}{5}$$The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none:
$$P(B) = 1 - frac{3}{5} frac{2}{4} frac{1}{3} = 1 - frac{1}{10} = frac{9}{10}$$There are ${5 choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find:
$$P(C) = frac{{5 choose 3}}{5 cdot 4 cdot 3} = frac{10}{60} = frac{1}{6}$$
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 11:49
jvdhooftjvdhooft
5,65961641
5,65961641
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%2f3096120%2fprobability-without-replacement-with-an-urn-with-five-balls-labeled-1-through-5%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
$begingroup$
There are $binom 53=10$ ways to choose $3$ balls out of the $5$, so your computation would yield $frac {11}{10}$ which is greater than $1$. Not sure I follow your computations for $(i)$ or $(ii)$.
$endgroup$
– lulu
Feb 1 at 11:42