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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?

The probability theory around a candy bagPicking balls from a bag until there are three of same colorWhat is the probability that each of the four candies selected is of a different type?There are 2 bags with 30 candies. One has 15 red/yellow and one has 10 red/10 yellow/10 purple. You pick a bag randomly, and pick 2 candies…Probability question candies 36 , 6 friendsWhat is the probability that if $5$ balls are distributed to $3$ bags (with no bag left empty) that there is exactly one ball in the first bag?What is the probability of choosing two candies that are not yellow?Bayes formula for candy corundumWhat is the probability of taking at least $2$ orange candies while taking $3$ candies from a bag with $6$ cherry, $3$ orange, and $2$ lemon candies?mean of buying different colored candy(tricky)

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

dantopa

6,53942244

asked 3 hours ago

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111

New contributor

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  $begingroup$
  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
  3 hours ago
  

  
  
  
  


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  100%. I'd purposely save a chocolate for last.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  


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  @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
  $endgroup$
  – TonyK
  3 hours ago
  
  
  

  
  
  
  


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  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
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

dantopa

6,53942244

asked 3 hours ago

The RangsterThe Rangster

111

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.

$endgroup$

• 
  
  
  

  
  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
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  100%. I'd purposely save a chocolate for last.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
  $endgroup$
  – TonyK
  3 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
  2 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

share|cite|improve this question

edited 3 hours ago

dantopa

6,53942244

asked 3 hours ago

The RangsterThe Rangster

111

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.

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share|cite|improve this question

edited 3 hours ago

dantopa

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asked 3 hours ago

The RangsterThe Rangster

111

New contributor

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edited 3 hours ago

dantopa

6,53942244

asked 3 hours ago

The RangsterThe Rangster

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edited 3 hours ago

dantopa

6,53942244

edited 3 hours ago

dantopa

6,53942244

asked 3 hours ago

The RangsterThe Rangster

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4 Answers
4

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4

$begingroup$

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 $dfrac{5}{8}timesdfrac{3}{7}$.

share|cite|improve this answer

answered 3 hours ago

TonyKTonyK

42.8k355134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a clever way of approaching this!
  $endgroup$
  – Remy
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 $binom{6}{2}/binom{8}{3} = 0.2679$.

share|cite|improve this answer

answered 3 hours ago

abcdabcd

1188

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

D.R.D.R.

1,514722

$endgroup$

add a comment | 

0

$begingroup$

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

$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$

share|cite|improve this answer

answered 1 hour ago

RemyRemy

6,504822

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4

$begingroup$

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 $dfrac{5}{8}timesdfrac{3}{7}$.

share|cite|improve this answer

answered 3 hours ago

TonyKTonyK

42.8k355134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a clever way of approaching this!
  $endgroup$
  – Remy
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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 $dfrac{5}{8}timesdfrac{3}{7}$.

share|cite|improve this answer

answered 3 hours ago

TonyKTonyK

42.8k355134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a clever way of approaching this!
  $endgroup$
  – Remy
  1 hour 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 $dfrac{5}{8}timesdfrac{3}{7}$.

share|cite|improve this answer

answered 3 hours ago

TonyKTonyK

42.8k355134

$endgroup$

share|cite|improve this answer

answered 3 hours ago

TonyKTonyK

42.8k355134

answered 3 hours ago

TonyKTonyK

42.8k355134

answered 3 hours ago

TonyKTonyK

42.8k355134

1

$begingroup$

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 $binom{6}{2}/binom{8}{3} = 0.2679$.

share|cite|improve this answer

answered 3 hours ago

abcdabcd

1188

$endgroup$

add a comment | 

1

$begingroup$

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 $binom{6}{2}/binom{8}{3} = 0.2679$.

share|cite|improve this answer

answered 3 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 $binom{6}{2}/binom{8}{3} = 0.2679$.

share|cite|improve this answer

answered 3 hours ago

abcdabcd

1188

$endgroup$

share|cite|improve this answer

answered 3 hours ago

abcdabcd

1188

answered 3 hours ago

abcdabcd

1188

answered 3 hours ago

abcdabcd

1188

0

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

D.R.D.R.

1,514722

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered 3 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.

share|cite|improve this answer

answered 3 hours ago

D.R.D.R.

1,514722

$endgroup$

share|cite|improve this answer

answered 3 hours ago

D.R.D.R.

1,514722

answered 3 hours ago

D.R.D.R.

1,514722

answered 3 hours ago

D.R.D.R.

1,514722

0

$begingroup$

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

$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$

share|cite|improve this answer

answered 1 hour ago

RemyRemy

6,504822

$endgroup$

add a comment | 

0

$begingroup$

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

$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$

share|cite|improve this answer

answered 1 hour ago

RemyRemy

6,504822

$endgroup$

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

$$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$

share|cite|improve this answer

answered 1 hour ago

RemyRemy

6,504822

$endgroup$

share|cite|improve this answer

answered 1 hour ago

RemyRemy

6,504822

answered 1 hour ago

RemyRemy

6,504822

answered 1 hour ago

RemyRemy

6,504822