Mixing
Given $5$ white balls, $8$ green balls and $7$ red balls. Find the probability of drawing a white ball then a...
Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.
What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.
probability combinatorics
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
add a comment |
Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.
What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.
probability combinatorics
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
add a comment |
Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.
What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.
probability combinatorics
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
Given $5$ white balls, $8$ green balls and $7$ red balls in an urn. Find out the probability to draw a white ball and then a green one if the drawing is done consecutively and after drawing the ball is returned into the urn.
What I ended up with as an answer is $1/20$ by taking the chance for drawing a white ball and multiplying it by the chance to get a green ball and then dividing by two, since I only want the case where the white ball is first, which I assume is half the cases.
probability combinatorics
probability combinatorics
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
edited Nov 21 '18 at 20:13
N. F. Taussig
43.6k93355
43.6k93355
asked Nov 21 '18 at 20:07
SartrSartr
949
asked Nov 21 '18 at 20:07
SartrSartr
949
asked Nov 21 '18 at 20:07
SartrSartr
949
949
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$
Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
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%2f3008279%2fgiven-5-white-balls-8-green-balls-and-7-red-balls-find-the-probability-o%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
There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
There is no need to divide by $2$. You need to multiply the probability of drawing a white ball on the first draw by the probability of drawing a green ball on the second draw. Since the draws are independent and the first ball is replaced, the probability of drawing a green ball on the second draw given that a white ball was drawn on the first draw is just the probability of drawing a green ball from the urn. Hence,
$$Pr(text{drawing white, then green}) = Pr(W)Pr(G mid W) = Pr(W)Pr(G) = frac{5}{20} cdot frac{8}{20} = frac{1}{10}$$
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 21 '18 at 20:18
N. F. TaussigN. F. Taussig
43.6k93355
43.6k93355
add a comment |
add a comment |
First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$
Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
add a comment |
First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$
Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
add a comment |
First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$
Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
First you calculate the probability of getting a white ball. $$P(white)=frac{Number of white balls}{Total number of balls} = frac{5}{20}$$
Then the probability of getting a green ball is $$P(green)=frac{Number of green balls}{Total number of balls} = frac{8}{20}$$
Because you put the ball back in the urn all the probabilities are independant
Thus $$P(White then Green) = P(White)*P_{White}(Green) = P(White)*P(Green) = frac{5*8}{20*20} = frac{1}{10}$$
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
answered Nov 21 '18 at 20:22
TheD0ubleTTheD0ubleT
39218
39218
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%2f3008279%2fgiven-5-white-balls-8-green-balls-and-7-red-balls-find-the-probability-o%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
