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Most likely value of k
Conditional Probability Problem (drawing chips from an urn)Probability Question: Drawing balls from an urnProbability: Pair of socks problemWhat is the probability of getting 2 same colour sweets and 1 different colour sweet?Probability of game ending with 10 quarters and 5 dimes?World Cup: what group stage result (vector) is most likely?probability of event given bag of five pairs of ballsThe last exam would be most likely occurring on?What is the % chance I will pick the correct random number in multiple drawingsHow many draws from an urn given c colors have been drawn?
$begingroup$
A while ago I got this question on my exam, anyone got an idea how to solve this?
Smarties are a chocolate candy that come in k different colors. Suppose that
we do not know k.
a.
We draw three smarties from a box and observe two different colors. What is the
most likely value for k0, given the smarties we have drawn.
Answer: 2
b.
We draw a fourth smartie and find a third color. What is now the most likely value
for k0, given the smarties we have drawn.
Answer: 5
probability probability-theory statistics statistical-inference
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A while ago I got this question on my exam, anyone got an idea how to solve this?
Smarties are a chocolate candy that come in k different colors. Suppose that
we do not know k.
a.
We draw three smarties from a box and observe two different colors. What is the
most likely value for k0, given the smarties we have drawn.
Answer: 2
b.
We draw a fourth smartie and find a third color. What is now the most likely value
for k0, given the smarties we have drawn.
Answer: 5
probability probability-theory statistics statistical-inference
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03
add a comment |
$begingroup$
A while ago I got this question on my exam, anyone got an idea how to solve this?
Smarties are a chocolate candy that come in k different colors. Suppose that
we do not know k.
a.
We draw three smarties from a box and observe two different colors. What is the
most likely value for k0, given the smarties we have drawn.
Answer: 2
b.
We draw a fourth smartie and find a third color. What is now the most likely value
for k0, given the smarties we have drawn.
Answer: 5
probability probability-theory statistics statistical-inference
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A while ago I got this question on my exam, anyone got an idea how to solve this?
Smarties are a chocolate candy that come in k different colors. Suppose that
we do not know k.
a.
We draw three smarties from a box and observe two different colors. What is the
most likely value for k0, given the smarties we have drawn.
Answer: 2
b.
We draw a fourth smartie and find a third color. What is now the most likely value
for k0, given the smarties we have drawn.
Answer: 5
probability probability-theory statistics statistical-inference
probability probability-theory statistics statistical-inference
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 9 at 17:58
J.DoeJ.Doe
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 9 at 17:58
J.DoeJ.Doe
82
asked Mar 9 at 17:58
J.DoeJ.Doe
82
82
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03
add a comment |
1
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03
1
1
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For part (a), the probability of drawing two different colours in a selection of $3$ is
$$binom{3}{1} k(k-1)(frac{1}{k})^3 = frac{3k(k-1)}{k^3}= frac{3}{k}-frac{3}{k^2}$$
The maximum value of this probability occurs when $frac{d}{dk} bigg(frac{3}{k}-frac{3}{k^2}bigg)=0 Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by
$$(frac{3}{k}-frac{3}{k^2})(frac{k-2}{k})=frac{3}{k}-frac{9}{k^2}+frac{6}{k^3}$$
The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
$endgroup$
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
add a comment |
Your Answer
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$begingroup$
For part (a), the probability of drawing two different colours in a selection of $3$ is
$$binom{3}{1} k(k-1)(frac{1}{k})^3 = frac{3k(k-1)}{k^3}= frac{3}{k}-frac{3}{k^2}$$
The maximum value of this probability occurs when $frac{d}{dk} bigg(frac{3}{k}-frac{3}{k^2}bigg)=0 Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by
$$(frac{3}{k}-frac{3}{k^2})(frac{k-2}{k})=frac{3}{k}-frac{9}{k^2}+frac{6}{k^3}$$
The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
$endgroup$
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
add a comment |
$begingroup$
For part (a), the probability of drawing two different colours in a selection of $3$ is
$$binom{3}{1} k(k-1)(frac{1}{k})^3 = frac{3k(k-1)}{k^3}= frac{3}{k}-frac{3}{k^2}$$
The maximum value of this probability occurs when $frac{d}{dk} bigg(frac{3}{k}-frac{3}{k^2}bigg)=0 Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by
$$(frac{3}{k}-frac{3}{k^2})(frac{k-2}{k})=frac{3}{k}-frac{9}{k^2}+frac{6}{k^3}$$
The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
$endgroup$
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
add a comment |
$begingroup$
For part (a), the probability of drawing two different colours in a selection of $3$ is
$$binom{3}{1} k(k-1)(frac{1}{k})^3 = frac{3k(k-1)}{k^3}= frac{3}{k}-frac{3}{k^2}$$
The maximum value of this probability occurs when $frac{d}{dk} bigg(frac{3}{k}-frac{3}{k^2}bigg)=0 Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by
$$(frac{3}{k}-frac{3}{k^2})(frac{k-2}{k})=frac{3}{k}-frac{9}{k^2}+frac{6}{k^3}$$
The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
$endgroup$
For part (a), the probability of drawing two different colours in a selection of $3$ is
$$binom{3}{1} k(k-1)(frac{1}{k})^3 = frac{3k(k-1)}{k^3}= frac{3}{k}-frac{3}{k^2}$$
The maximum value of this probability occurs when $frac{d}{dk} bigg(frac{3}{k}-frac{3}{k^2}bigg)=0 Rightarrow k=2$ and hence the most likely value of $k$ is $2$.
For part (b), the probability of finding a third different colour after drawing $3$ in part (a) is given by
$$(frac{3}{k}-frac{3}{k^2})(frac{k-2}{k})=frac{3}{k}-frac{9}{k^2}+frac{6}{k^3}$$
The maximum value of this probability occurs when $k=5$ and hence the most likely value of $k$ is $5$.
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
edited Mar 9 at 18:59
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
answered Mar 9 at 18:30
Peter ForemanPeter Foreman
3,7861216
3,7861216
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
add a comment |
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
In part a), the probability is twice what you give, because there are two ways to assign the selected colors to the candies.
$endgroup$
– saulspatz
Mar 9 at 18:50
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
Thanks, I forgot that either permutation was valid.
$endgroup$
– Peter Foreman
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
I'm sure this is the answer the examiner wanted, but it bothers me. If "most likely" means "with highest probability" how are we to assign probabilities to $k?$ In part a), $$sum_{k=2}^inftyleft(frac3k-{3over k^2}right)=infty$$ so we can't just normalize them. Perhaps I'm becoming a Bayesian in my old age.
$endgroup$
– saulspatz
Mar 9 at 19:00
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@saulspatz - see e.g. stats.stackexchange.com/questions/2641/… but the short short answer is each term in your summation is a prob value in a different experiement (different universe?) where the value of $k$ is different, so you can't add them. This used to confuse me too... if that makes you feel better. :) Also, re: Bayesian, if you can imagine a magical prior distribution on all $k$ with equal probability (magical coz it doesn't exist), then that is about equivalent to the max-likelihood viewpoint.
$endgroup$
– antkam
Mar 9 at 19:59
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
$begingroup$
@antkam Thanks you. The accepted answer does clear it up for me.
$endgroup$
– saulspatz
Mar 9 at 20:05
add a comment |
J.Doe is a new contributor. Be nice, and check out our Code of Conduct.
J.Doe is a new contributor. Be nice, and check out our Code of Conduct.
J.Doe is a new contributor. Be nice, and check out our Code of Conduct.
J.Doe is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
I think you need to impose a prior probability distribution on k. Maybe an exponential distribution? Then you can apply bayes rule. But the question does not have a correct answer as it is given.
$endgroup$
– Mark
Mar 9 at 18:03