There is a bag of 8 candies, and 3 are chocolates. You eat candy until the chocolates are gone. What is the...
$begingroup$
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 Feb 26 at 0:40
dantopa
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
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closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54
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 Feb 26 at 0:40
dantopa
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
$endgroup$
closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54
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.
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– dantopa
Feb 26 at 0:40
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
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
Feb 26 at 1:13
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 Feb 26 at 0:40
dantopa
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
$endgroup$
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 Feb 26 at 0:40
dantopa
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
edited Feb 26 at 0:40
dantopa
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
edited Feb 26 at 0:40
dantopa
6,64442245
edited Feb 26 at 0:40
dantopa
6,64442245
edited Feb 26 at 0:40
dantopa
6,64442245
6,64442245
asked Feb 26 at 0:36
The RangsterThe Rangster
161
asked Feb 26 at 0:36
The RangsterThe Rangster
161
asked Feb 26 at 0:36
The RangsterThe Rangster
161
161
closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54
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.
closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54
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.
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– dantopa
Feb 26 at 0:40
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
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
Feb 26 at 1:13
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
Feb 26 at 0:40
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
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
Feb 26 at 1:13
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
Feb 26 at 0:40
$begingroup$
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
Feb 26 at 0:40
1
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
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
Feb 26 at 1:13
$begingroup$
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
Feb 26 at 1:13
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 Feb 26 at 1:07
TonyKTonyK
43.7k358137
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
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 Feb 26 at 0:54
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 Feb 26 at 2:52
RemyRemy
6,642922
$endgroup$
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 Feb 26 at 0:45
D.R.D.R.
1,776823
$endgroup$
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 Feb 26 at 1:07
TonyKTonyK
43.7k358137
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
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 Feb 26 at 1:07
TonyKTonyK
43.7k358137
$endgroup$
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
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 Feb 26 at 1:07
TonyKTonyK
43.7k358137
$endgroup$
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 Feb 26 at 1:07
TonyKTonyK
43.7k358137
answered Feb 26 at 1:07
TonyKTonyK
43.7k358137
answered Feb 26 at 1:07
TonyKTonyK
43.7k358137
answered Feb 26 at 1:07
TonyKTonyK
43.7k358137
43.7k358137
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
add a comment |
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
$begingroup$
This is a clever way of approaching this!
$endgroup$
– Remy
Feb 26 at 2:48
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 Feb 26 at 0:54
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 Feb 26 at 0:54
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 Feb 26 at 0:54
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 Feb 26 at 0:54
abcdabcd
1188
answered Feb 26 at 0:54
abcdabcd
1188
answered Feb 26 at 0:54
abcdabcd
1188
answered Feb 26 at 0:54
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 Feb 26 at 2:52
RemyRemy
6,642922
$endgroup$
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 Feb 26 at 2:52
RemyRemy
6,642922
$endgroup$
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 Feb 26 at 2:52
RemyRemy
6,642922
$endgroup$
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 Feb 26 at 2:52
RemyRemy
6,642922
answered Feb 26 at 2:52
RemyRemy
6,642922
answered Feb 26 at 2:52
RemyRemy
6,642922
answered Feb 26 at 2:52
RemyRemy
6,642922
6,642922
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 Feb 26 at 0:45
D.R.D.R.
1,776823
$endgroup$
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 Feb 26 at 0:45
D.R.D.R.
1,776823
$endgroup$
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 Feb 26 at 0:45
D.R.D.R.
1,776823
$endgroup$
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 Feb 26 at 0:45
D.R.D.R.
1,776823
answered Feb 26 at 0:45
D.R.D.R.
1,776823
answered Feb 26 at 0:45
D.R.D.R.
1,776823
answered Feb 26 at 0:45
D.R.D.R.
1,776823
1,776823
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
Feb 26 at 0:40
1
$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53
$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08
$begingroup$
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
Feb 26 at 1:13