Mixing
Expected Value Of Number of Removals From Urn
$begingroup$
Here's the question: "An urn contains b blue balls and r red balls. You repeatedly and independently remove balls from urn (without returning them) until the first blue ball is drawn. All balls currently in the urn have an equal probability of being selected each time a ball is removed. Define the random variable X as the number of balls that are drawn (number of red balls that are removed plus the first blue ball that is removed). Find E[X], the expected value of random variable X."
I'm having trouble finding a closed form expression for the expected value. Here's the work I've done:
probability statistics
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
$endgroup$
|
show 4 more comments
$begingroup$
Here's the question: "An urn contains b blue balls and r red balls. You repeatedly and independently remove balls from urn (without returning them) until the first blue ball is drawn. All balls currently in the urn have an equal probability of being selected each time a ball is removed. Define the random variable X as the number of balls that are drawn (number of red balls that are removed plus the first blue ball that is removed). Find E[X], the expected value of random variable X."
I'm having trouble finding a closed form expression for the expected value. Here's the work I've done:
probability statistics
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
$endgroup$
$begingroup$
@callculus: no, the balls aren't replaced.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:08
$begingroup$
@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
$endgroup$
– callculus
Mar 3 '17 at 23:13
$begingroup$
What is meant by define $x$?
$endgroup$
– VortexYT
Mar 3 '17 at 23:13
$begingroup$
@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:16
$begingroup$
Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
$endgroup$
– LeBron James
Mar 3 '17 at 23:17
|
show 4 more comments
$begingroup$
Here's the question: "An urn contains b blue balls and r red balls. You repeatedly and independently remove balls from urn (without returning them) until the first blue ball is drawn. All balls currently in the urn have an equal probability of being selected each time a ball is removed. Define the random variable X as the number of balls that are drawn (number of red balls that are removed plus the first blue ball that is removed). Find E[X], the expected value of random variable X."
I'm having trouble finding a closed form expression for the expected value. Here's the work I've done:
probability statistics
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
$endgroup$
Here's the question: "An urn contains b blue balls and r red balls. You repeatedly and independently remove balls from urn (without returning them) until the first blue ball is drawn. All balls currently in the urn have an equal probability of being selected each time a ball is removed. Define the random variable X as the number of balls that are drawn (number of red balls that are removed plus the first blue ball that is removed). Find E[X], the expected value of random variable X."
I'm having trouble finding a closed form expression for the expected value. Here's the work I've done:
probability statistics
probability statistics
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
edited Mar 4 '17 at 7:30
Graham Kemp
87.8k43578
87.8k43578
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
asked Mar 4 '17 at 1:09
LeBron JamesLeBron James
213
213
$begingroup$
@callculus: no, the balls aren't replaced.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:08
$begingroup$
@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
$endgroup$
– callculus
Mar 3 '17 at 23:13
$begingroup$
What is meant by define $x$?
$endgroup$
– VortexYT
Mar 3 '17 at 23:13
$begingroup$
@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:16
$begingroup$
Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
$endgroup$
– LeBron James
Mar 3 '17 at 23:17
|
show 4 more comments
$begingroup$
@callculus: no, the balls aren't replaced.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:08
$begingroup$
@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
$endgroup$
– callculus
Mar 3 '17 at 23:13
$begingroup$
What is meant by define $x$?
$endgroup$
– VortexYT
Mar 3 '17 at 23:13
$begingroup$
@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:16
$begingroup$
Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
$endgroup$
– LeBron James
Mar 3 '17 at 23:17
$begingroup$
@callculus: no, the balls aren't replaced.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:08
$begingroup$
@callculus: no, the balls aren't replaced.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:08
$begingroup$
@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
$endgroup$
– callculus
Mar 3 '17 at 23:13
$begingroup$
@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
$endgroup$
– callculus
Mar 3 '17 at 23:13
$begingroup$
What is meant by define $x$?
$endgroup$
– VortexYT
Mar 3 '17 at 23:13
$begingroup$
What is meant by define $x$?
$endgroup$
– VortexYT
Mar 3 '17 at 23:13
$begingroup$
@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:16
$begingroup$
@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
$endgroup$
– Yves Daoust
Mar 3 '17 at 23:16
$begingroup$
Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
$endgroup$
– LeBron James
Mar 3 '17 at 23:17
$begingroup$
Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
$endgroup$
– LeBron James
Mar 3 '17 at 23:17
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
The first drawing gives a blue with probability $d_1=b/(b+r)$.
Otherwise, the second drawing gives a blue with conditional probability $d_2=b/(b+r-1)$.
Otherwise, the third drawing gives a blue with conditional probability $d_3=b/(b+r-2)$.
...
Otherwise, the $r+1^{th}$ drawing gives a blue with conditional probability $d_{r+1}=b/b$.
Then
$$p_1=d_1,\
p_2=(1-p_1)d_2=d_2-d_1d_2,\
p_3=(1-p_2)d_3=d_3-d_2d_3+d_1d_2d_3,\
cdots\
p_{r+1}=(1-p_r)d_{r+1}=d_{r+1}-d_rd_{r+1}+d_{r-1}d_rd_{r+1}+cdots pm d_1d_2d_3cdots d_{r+1}.$$
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
Let $E[r,b]$ denote the desired answer.
Consider the problem this way:
You have $r$ red balls in a line and you are going to insert $b$ blues into that line. These will divide the red balls into $b+1$ strings of red balls. By symmetry, each of these is expected to have the same length which must then be $frac {r}{b+1}$. It follows that your answer is $$E[r,b]=boxed {frac r{b+1}+1}$$
Note 1: to get intuition for the symmetry argument, first imagine that the red balls are arranged in a circle instead of a line and that there is $1$ yellow ball added in with the blues. It is now clear I think that the gaps must have the same expected length $frac r{b+1}$ and deleting the yellow ball restores the problem to the original.
Note 2: If you still find the symmetry argument hard to intuit, the answer can be verified by looking at a recursion. The first draw is either red or blue, so we get $$E[r,b]=frac b{r+b}times 1+frac r{r+b}times left(E[r-1,b]+1right)$$ and it is straightforward to verify that the answer obtained above satisfies this (along with the obvious boundary condition $E[0,b]=1$ for $b>0$).
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
$endgroup$
$begingroup$
How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
The first drawing gives a blue with probability $d_1=b/(b+r)$.
Otherwise, the second drawing gives a blue with conditional probability $d_2=b/(b+r-1)$.
Otherwise, the third drawing gives a blue with conditional probability $d_3=b/(b+r-2)$.
...
Otherwise, the $r+1^{th}$ drawing gives a blue with conditional probability $d_{r+1}=b/b$.
Then
$$p_1=d_1,\
p_2=(1-p_1)d_2=d_2-d_1d_2,\
p_3=(1-p_2)d_3=d_3-d_2d_3+d_1d_2d_3,\
cdots\
p_{r+1}=(1-p_r)d_{r+1}=d_{r+1}-d_rd_{r+1}+d_{r-1}d_rd_{r+1}+cdots pm d_1d_2d_3cdots d_{r+1}.$$
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
Hint:
The first drawing gives a blue with probability $d_1=b/(b+r)$.
Otherwise, the second drawing gives a blue with conditional probability $d_2=b/(b+r-1)$.
Otherwise, the third drawing gives a blue with conditional probability $d_3=b/(b+r-2)$.
...
Otherwise, the $r+1^{th}$ drawing gives a blue with conditional probability $d_{r+1}=b/b$.
Then
$$p_1=d_1,\
p_2=(1-p_1)d_2=d_2-d_1d_2,\
p_3=(1-p_2)d_3=d_3-d_2d_3+d_1d_2d_3,\
cdots\
p_{r+1}=(1-p_r)d_{r+1}=d_{r+1}-d_rd_{r+1}+d_{r-1}d_rd_{r+1}+cdots pm d_1d_2d_3cdots d_{r+1}.$$
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
Hint:
The first drawing gives a blue with probability $d_1=b/(b+r)$.
Otherwise, the second drawing gives a blue with conditional probability $d_2=b/(b+r-1)$.
Otherwise, the third drawing gives a blue with conditional probability $d_3=b/(b+r-2)$.
...
Otherwise, the $r+1^{th}$ drawing gives a blue with conditional probability $d_{r+1}=b/b$.
Then
$$p_1=d_1,\
p_2=(1-p_1)d_2=d_2-d_1d_2,\
p_3=(1-p_2)d_3=d_3-d_2d_3+d_1d_2d_3,\
cdots\
p_{r+1}=(1-p_r)d_{r+1}=d_{r+1}-d_rd_{r+1}+d_{r-1}d_rd_{r+1}+cdots pm d_1d_2d_3cdots d_{r+1}.$$
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
$endgroup$
Hint:
The first drawing gives a blue with probability $d_1=b/(b+r)$.
Otherwise, the second drawing gives a blue with conditional probability $d_2=b/(b+r-1)$.
Otherwise, the third drawing gives a blue with conditional probability $d_3=b/(b+r-2)$.
...
Otherwise, the $r+1^{th}$ drawing gives a blue with conditional probability $d_{r+1}=b/b$.
Then
$$p_1=d_1,\
p_2=(1-p_1)d_2=d_2-d_1d_2,\
p_3=(1-p_2)d_3=d_3-d_2d_3+d_1d_2d_3,\
cdots\
p_{r+1}=(1-p_r)d_{r+1}=d_{r+1}-d_rd_{r+1}+d_{r-1}d_rd_{r+1}+cdots pm d_1d_2d_3cdots d_{r+1}.$$
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
edited Mar 3 '17 at 23:29
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
answered Mar 3 '17 at 23:07
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
$begingroup$
Let $E[r,b]$ denote the desired answer.
Consider the problem this way:
You have $r$ red balls in a line and you are going to insert $b$ blues into that line. These will divide the red balls into $b+1$ strings of red balls. By symmetry, each of these is expected to have the same length which must then be $frac {r}{b+1}$. It follows that your answer is $$E[r,b]=boxed {frac r{b+1}+1}$$
Note 1: to get intuition for the symmetry argument, first imagine that the red balls are arranged in a circle instead of a line and that there is $1$ yellow ball added in with the blues. It is now clear I think that the gaps must have the same expected length $frac r{b+1}$ and deleting the yellow ball restores the problem to the original.
Note 2: If you still find the symmetry argument hard to intuit, the answer can be verified by looking at a recursion. The first draw is either red or blue, so we get $$E[r,b]=frac b{r+b}times 1+frac r{r+b}times left(E[r-1,b]+1right)$$ and it is straightforward to verify that the answer obtained above satisfies this (along with the obvious boundary condition $E[0,b]=1$ for $b>0$).
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
$endgroup$
$begingroup$
How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
|
show 2 more comments
$begingroup$
Let $E[r,b]$ denote the desired answer.
Consider the problem this way:
You have $r$ red balls in a line and you are going to insert $b$ blues into that line. These will divide the red balls into $b+1$ strings of red balls. By symmetry, each of these is expected to have the same length which must then be $frac {r}{b+1}$. It follows that your answer is $$E[r,b]=boxed {frac r{b+1}+1}$$
Note 1: to get intuition for the symmetry argument, first imagine that the red balls are arranged in a circle instead of a line and that there is $1$ yellow ball added in with the blues. It is now clear I think that the gaps must have the same expected length $frac r{b+1}$ and deleting the yellow ball restores the problem to the original.
Note 2: If you still find the symmetry argument hard to intuit, the answer can be verified by looking at a recursion. The first draw is either red or blue, so we get $$E[r,b]=frac b{r+b}times 1+frac r{r+b}times left(E[r-1,b]+1right)$$ and it is straightforward to verify that the answer obtained above satisfies this (along with the obvious boundary condition $E[0,b]=1$ for $b>0$).
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
$endgroup$
$begingroup$
How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
|
show 2 more comments
$begingroup$
Let $E[r,b]$ denote the desired answer.
Consider the problem this way:
You have $r$ red balls in a line and you are going to insert $b$ blues into that line. These will divide the red balls into $b+1$ strings of red balls. By symmetry, each of these is expected to have the same length which must then be $frac {r}{b+1}$. It follows that your answer is $$E[r,b]=boxed {frac r{b+1}+1}$$
Note 1: to get intuition for the symmetry argument, first imagine that the red balls are arranged in a circle instead of a line and that there is $1$ yellow ball added in with the blues. It is now clear I think that the gaps must have the same expected length $frac r{b+1}$ and deleting the yellow ball restores the problem to the original.
Note 2: If you still find the symmetry argument hard to intuit, the answer can be verified by looking at a recursion. The first draw is either red or blue, so we get $$E[r,b]=frac b{r+b}times 1+frac r{r+b}times left(E[r-1,b]+1right)$$ and it is straightforward to verify that the answer obtained above satisfies this (along with the obvious boundary condition $E[0,b]=1$ for $b>0$).
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
$endgroup$
Let $E[r,b]$ denote the desired answer.
Consider the problem this way:
You have $r$ red balls in a line and you are going to insert $b$ blues into that line. These will divide the red balls into $b+1$ strings of red balls. By symmetry, each of these is expected to have the same length which must then be $frac {r}{b+1}$. It follows that your answer is $$E[r,b]=boxed {frac r{b+1}+1}$$
Note 1: to get intuition for the symmetry argument, first imagine that the red balls are arranged in a circle instead of a line and that there is $1$ yellow ball added in with the blues. It is now clear I think that the gaps must have the same expected length $frac r{b+1}$ and deleting the yellow ball restores the problem to the original.
Note 2: If you still find the symmetry argument hard to intuit, the answer can be verified by looking at a recursion. The first draw is either red or blue, so we get $$E[r,b]=frac b{r+b}times 1+frac r{r+b}times left(E[r-1,b]+1right)$$ and it is straightforward to verify that the answer obtained above satisfies this (along with the obvious boundary condition $E[0,b]=1$ for $b>0$).
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
edited Mar 4 '17 at 1:44
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
answered Mar 4 '17 at 1:30
lulululu
43.6k25081
43.6k25081
$begingroup$
How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
|
show 2 more comments
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How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
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– LeBron James
Mar 4 '17 at 2:10
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Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
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– LeBron James
Mar 4 '17 at 2:14
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Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
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– lulu
Mar 4 '17 at 2:15
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Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
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– LeBron James
Mar 4 '17 at 2:18
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That's just the definition of the Expected Value. Of course the recursion uses the definition.
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– lulu
Mar 4 '17 at 2:19
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How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
How do you solve this problem without using the symmetry argument, but with the argument I have used in my work?
$endgroup$
– LeBron James
Mar 4 '17 at 2:10
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Can you solve the problem using algebra and probabilities instead of symmetry/intuition arguments? Look at the work I've done.
$endgroup$
– LeBron James
Mar 4 '17 at 2:14
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Well, I wouldn't. I don't even see a quick way to solve the recursion I wrote down (though surely there must be one). Manipulating identities between all the binomial coefficients may well be possible, but likely to be hard (and the proofs of those identities will likely involve symmetry). But the recursive argument (effectively proof by induction on $r$) is purely algebraic.
$endgroup$
– lulu
Mar 4 '17 at 2:15
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
Can you walk through how to do this algebraically with summations? Please look at the work I've posted. In addition, we should use these identities to help arrive at the answer: i.imgur.com/4amu4EU.png, i.imgur.com/ICLFw4N.png
$endgroup$
– LeBron James
Mar 4 '17 at 2:18
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
$begingroup$
That's just the definition of the Expected Value. Of course the recursion uses the definition.
$endgroup$
– lulu
Mar 4 '17 at 2:19
|
show 2 more comments
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@callculus: no, the balls aren't replaced.
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– Yves Daoust
Mar 3 '17 at 23:08
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@YvesDaoust "All balls have an equal probability of being selected each time a ball is removed." Due to this part I thought it is with replacement, especially the part "equal probability".
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– callculus
Mar 3 '17 at 23:13
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What is meant by define $x$?
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– VortexYT
Mar 3 '17 at 23:13
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@callculus: the term "removed" is used twice. The term "drawn" is also used twice, but only for the last ball.
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– Yves Daoust
Mar 3 '17 at 23:16
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Balls are removed and not replaced. You stop taking out balls until you remove your first blue ball, and the random variable X takes the value for how many draws it took to get to this point.
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– LeBron James
Mar 3 '17 at 23:17