There is a bag of 8 candies, and 3 are chocolates. You eat candy until the chocolates are gone. What is the...
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There is a bag of 8 candies, and 3 are chocolates. You eat candy until the chocolates are gone. What is the probability you will have eaten 7 candies? [on hold]
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$begingroup$
You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?
probability statistics
edited 6 hours ago
dantopa
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?
probability statistics
edited 6 hours ago
dantopa
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago
add a comment |
$begingroup$
You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?
probability statistics
edited 6 hours ago
dantopa
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?
probability statistics
probability statistics
edited 6 hours ago
dantopa
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
dantopa
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
dantopa
6,53942244
edited 6 hours ago
dantopa
6,53942244
edited 6 hours ago
dantopa
6,53942244
6,53942244
asked 6 hours ago
The RangsterThe Rangster
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
The RangsterThe Rangster
161
asked 6 hours ago
The RangsterThe Rangster
161
161
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago
add a comment |
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago
1
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
1
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.
answered 5 hours ago
TonyKTonyK
42.8k355134
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
add a comment |
$begingroup$
As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.
answered 5 hours ago
abcdabcd
1188
$endgroup$
add a comment |
$begingroup$
As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is
$$frac{{3choose2} {5choose4}}{8choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is
$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$
answered 3 hours ago
RemyRemy
6,514822
$endgroup$
add a comment |
$begingroup$
HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
answered 6 hours ago
D.R.D.R.
1,514722
$endgroup$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.
answered 5 hours ago
TonyKTonyK
42.8k355134
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
add a comment |
$begingroup$
A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.
answered 5 hours ago
TonyKTonyK
42.8k355134
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
add a comment |
$begingroup$
A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.
answered 5 hours ago
TonyKTonyK
42.8k355134
$endgroup$
A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:
What is the probability that the first pick is not a chocolate, and the second pick is?
So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.
answered 5 hours ago
TonyKTonyK
42.8k355134
answered 5 hours ago
TonyKTonyK
42.8k355134
answered 5 hours ago
TonyKTonyK
42.8k355134
answered 5 hours ago
TonyKTonyK
42.8k355134
42.8k355134
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
add a comment |
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
3 hours ago
add a comment |
$begingroup$
As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.
answered 5 hours ago
abcdabcd
1188
$endgroup$
add a comment |
$begingroup$
As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.
answered 5 hours ago
abcdabcd
1188
$endgroup$
add a comment |
$begingroup$
As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.
answered 5 hours ago
abcdabcd
1188
$endgroup$
As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.
answered 5 hours ago
abcdabcd
1188
answered 5 hours ago
abcdabcd
1188
answered 5 hours ago
abcdabcd
1188
answered 5 hours ago
abcdabcd
1188
1188
add a comment |
add a comment |
$begingroup$
As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is
$$frac{{3choose2} {5choose4}}{8choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is
$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$
answered 3 hours ago
RemyRemy
6,514822
$endgroup$
add a comment |
$begingroup$
As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is
$$frac{{3choose2} {5choose4}}{8choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is
$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$
answered 3 hours ago
RemyRemy
6,514822
$endgroup$
add a comment |
$begingroup$
As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is
$$frac{{3choose2} {5choose4}}{8choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is
$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$
answered 3 hours ago
RemyRemy
6,514822
$endgroup$
As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is
$$frac{{3choose2} {5choose4}}{8choose6}$$
This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is
$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$
answered 3 hours ago
RemyRemy
6,514822
answered 3 hours ago
RemyRemy
6,514822
answered 3 hours ago
RemyRemy
6,514822
answered 3 hours ago
RemyRemy
6,514822
6,514822
add a comment |
add a comment |
$begingroup$
HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
answered 6 hours ago
D.R.D.R.
1,514722
$endgroup$
add a comment |
$begingroup$
HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
answered 6 hours ago
D.R.D.R.
1,514722
$endgroup$
add a comment |
$begingroup$
HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
answered 6 hours ago
D.R.D.R.
1,514722
$endgroup$
HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.
If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.
answered 6 hours ago
D.R.D.R.
1,514722
answered 6 hours ago
D.R.D.R.
1,514722
answered 6 hours ago
D.R.D.R.
1,514722
answered 6 hours ago
D.R.D.R.
1,514722
1,514722
add a comment |
add a comment |
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
6 hours ago
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
5 hours ago
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
5 hours ago
$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
5 hours ago